Integral of the form ex(F(x) + F'(x)) dx Questions in English

(B) We know that $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c$.
Given integral is $I = \int e^x \left( \frac{1 + \sin x}{1 + \cos x} \right) dx$.
Using trigonometric identities: $1 + \sin x = 1 + 2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $1 + \cos x = 2 \cos^2 \frac{x}{2}$.
Substituting these into the expression: $\frac{1 + \sin x}{1 + \cos x} = \frac{1 + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} = \frac{1}{2 \cos^2 \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} = \frac{1}{2} \sec^2 \frac{x}{2} + \tan \frac{x}{2}$.
Let $f(x) = \tan \frac{x}{2}$,then $f'(x) = \sec^2 \frac{x}{2} \cdot \frac{1}{2} = \frac{1}{2} \sec^2 \frac{x}{2}$.
Thus,the integral becomes $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c = e^x \tan \frac{x}{2} + c$.

(B) Let $I = \int \left\{ \cos^{-1} x - (1-x^2)^{-\frac{1}{2}} \right\} k \, dx$.
Given that $\int \left\{ \cos^{-1} x - \frac{1}{\sqrt{1-x^2}} \right\} k \, dx = k \cdot \cos^{-1} x + c$.
Let $f(x) = \cos^{-1} x$. Then $f'(x) = -\frac{1}{\sqrt{1-x^2}}$.
The integral becomes $\int k \left\{ f(x) + f'(x) \right\} dx = k \cdot \cos^{-1} x + c$.
We know that $\int e^x \{ f(x) + f'(x) \} dx = e^x f(x) + c$.
Comparing this with the given equation $\int k \{ f(x) + f'(x) \} dx = k \cdot f(x) + c$,we can see that $k$ must be a function of $x$ such that $k = e^x$.
Thus,$k = e^x$.

(C) We know that $1 + \tan^2 x = \sec^2 x$.
Substituting this into the integral,we get:
$\int e^x(2021 + \tan x + \tan^2 x) dx = \int e^x(2020 + 1 + \tan^2 x + \tan x) dx$
$= \int e^x(2020 + \sec^2 x + \tan x) dx$
$= \int e^x(2020 + \tan x) dx + \int e^x \sec^2 x dx$.
Let $f(x) = 2020 + \tan x$. Then $f'(x) = \sec^2 x$.
Using the standard integral formula $\int e^x(f(x) + f'(x)) dx = e^x f(x) + C$,we have:
$\int e^x(2020 + \tan x + \sec^2 x) dx = e^x(2020 + \tan x) + C$.
Thus,the correct option is $C$.

(B) We use the standard integral formula: $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$.
Given the integral: $I = \int \frac{x+100}{(x+101)^2} e^x \, dx$.
We can rewrite the numerator as: $x + 100 = (x + 101) - 1$.
So,the integral becomes: $I = \int \frac{(x+101) - 1}{(x+101)^2} e^x \, dx$.
$I = \int \left( \frac{x+101}{(x+101)^2} - \frac{1}{(x+101)^2} \right) e^x \, dx$.
$I = \int \left( \frac{1}{x+101} - \frac{1}{(x+101)^2} \right) e^x \, dx$.
Let $f(x) = \frac{1}{x+101}$.
Then $f'(x) = -\frac{1}{(x+101)^2}$.
Since the expression is in the form $\int e^x [f(x) + f'(x)] \, dx$,the result is $e^x f(x) + C$.
Therefore,$I = e^x \left( \frac{1}{x+101} \right) + C = \frac{e^x}{x+101} + C$.

(B) We know that the integral of the form $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$.
Given the integral is $\int e^x \cdot \sec x(1 + \tan x) \, dx$.
This can be rewritten as $\int e^x (\sec x + \sec x \tan x) \, dx$.
Here,let $f(x) = \sec x$.
Then,the derivative $f'(x) = \sec x \tan x$.
Substituting these into the formula,we get $\int e^x (f(x) + f'(x)) \, dx = e^x \sec x + C$.
Therefore,the correct option is $B$.

(A) We need to evaluate the integral $I = \int e^x (\tan x + \tan^2 x) \, dx$.
Recall the identity $1 + \tan^2 x = \sec^2 x$,which implies $\tan^2 x = \sec^2 x - 1$.
Substituting this into the integral,we get:
$I = \int e^x (\tan x + \sec^2 x - 1) \, dx$.
Rearranging the terms:
$I = \int e^x ((\tan x - 1) + \sec^2 x) \, dx$.
Let $f(x) = \tan x - 1$.
Then,$f'(x) = \sec^2 x$.
Using the standard integration formula $\int e^x (f(x) + f'(x)) \, dx = e^x f(x) + C$,we have:
$I = e^x (\tan x - 1) + C$.
Therefore,the correct option is $A$.

(D) We want to evaluate the integral $I = \int \frac{(x-3) e^x}{(x-1)^3} d x$.
Rewrite the numerator as $(x-1-2)$:
$I = \int \frac{(x-1-2) e^x}{(x-1)^3} d x$
$I = \int \left( \frac{x-1}{(x-1)^3} - \frac{2}{(x-1)^3} \right) e^x d x$
$I = \int \left( \frac{1}{(x-1)^2} - \frac{2}{(x-1)^3} \right) e^x d x$
Recall the standard integral form $\int e^x [f(x) + f'(x)] d x = e^x f(x) + C$.
Let $f(x) = \frac{1}{(x-1)^2} = (x-1)^{-2}$.
Then $f'(x) = -2(x-1)^{-3} = -\frac{2}{(x-1)^3}$.
Since the integrand is in the form $e^x [f(x) + f'(x)]$,the integral is $e^x f(x) + C$.
Therefore,$I = \frac{e^x}{(x-1)^2} + C$.

(A) We know that $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + c$.
First,rewrite the integrand: $\frac{x^2+1}{(x+1)^2} = \frac{x^2-1+2}{(x+1)^2} = \frac{(x-1)(x+1)+2}{(x+1)^2} = \frac{x-1}{x+1} + \frac{2}{(x+1)^2}$.
Let $f(x) = \frac{x-1}{x+1}$.
Then $f'(x) = \frac{(x+1)(1) - (x-1)(1)}{(x+1)^2} = \frac{x+1-x+1}{(x+1)^2} = \frac{2}{(x+1)^2}$.
Thus,the integral becomes $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + c = \left( \frac{x-1}{x+1} \right) e^x + c$.

(A) Given the integral $ I = \int e^{x} \left( \frac{1+\sin x}{1+\cos x} \right) dx $.
Using trigonometric identities $ 1+\cos x = 2\cos^2 \left( \frac{x}{2} \right) $ and $ \sin x = 2\sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right) $:
$ I = \int e^{x} \left( \frac{1 + 2\sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right)}{2\cos^2 \left( \frac{x}{2} \right)} \right) dx $
$ I = \int e^{x} \left( \frac{1}{2\cos^2 \left( \frac{x}{2} \right)} + \frac{2\sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right)}{2\cos^2 \left( \frac{x}{2} \right)} \right) dx $
$ I = \int e^{x} \left( \frac{1}{2} \sec^2 \left( \frac{x}{2} \right) + \tan \left( \frac{x}{2} \right) \right) dx $
We know that $ \int e^{x} (f(x) + f'(x)) dx = e^{x} f(x) + C $.
Here,let $ f(x) = \tan \left( \frac{x}{2} \right) $. Then $ f'(x) = \sec^2 \left( \frac{x}{2} \right) \cdot \frac{1}{2} = \frac{1}{2} \sec^2 \left( \frac{x}{2} \right) $.
Thus,$ I = e^{x} \tan \left( \frac{x}{2} \right) + C $.

(D) Let $I = \int \frac{x e^{x} d x}{(1+x)^{2}}$.
We can rewrite the numerator as $(x+1-1)$.
$I = \int \frac{e^{x}(x+1-1)}{(1+x)^{2}} d x$.
$I = \int e^{x} \left[ \frac{x+1}{(1+x)^{2}} - \frac{1}{(1+x)^{2}} \right] d x$.
$I = \int e^{x} \left[ \frac{1}{1+x} - \frac{1}{(1+x)^{2}} \right] d x$.
Let $f(x) = \frac{1}{1+x}$. Then $f'(x) = -\frac{1}{(1+x)^{2}}$.
Using the standard integration formula $\int e^{x} [f(x) + f'(x)] d x = e^{x} f(x) + C$,we get:
$I = e^{x} \left( \frac{1}{1+x} \right) + C = \frac{e^{x}}{1+x} + C$.

(A) Let $I = \int e^{x} \left( \frac{1+\sin x}{1+\cos x} \right) dx$.
Using trigonometric identities,$1+\sin x = 1+2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $1+\cos x = 2 \cos^2 \frac{x}{2}$.
Substituting these into the integral:
$I = \int e^{x} \left( \frac{1+2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \right) dx$
$I = \int e^{x} \left( \frac{1}{2 \cos^2 \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \right) dx$
$I = \int e^{x} \left( \frac{1}{2} \sec^2 \frac{x}{2} + \tan \frac{x}{2} \right) dx$.
Let $f(x) = \tan \frac{x}{2}$. Then $f'(x) = \sec^2 \frac{x}{2} \cdot \frac{1}{2} = \frac{1}{2} \sec^2 \frac{x}{2}$.
Since the integral is in the form $\int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + C$,
we get $I = e^{x} \tan \frac{x}{2} + C$.

(A) Given that,$ I = \int \frac{e^{x}(x^{2} \tan^{-1} x + \tan^{-1} x + 1)}{x^{2} + 1} dx $.
We can rewrite the numerator by grouping terms:
$ I = \int e^{x} \left( \frac{(x^{2} + 1) \tan^{-1} x + 1}{x^{2} + 1} \right) dx $.
Dividing each term by $ x^{2} + 1 $:
$ I = \int e^{x} \left( \tan^{-1} x + \frac{1}{x^{2} + 1} \right) dx $.
Let $ f(x) = \tan^{-1} x $. Then $ f'(x) = \frac{1}{x^{2} + 1} $.
Using the standard integral formula $ \int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + c $:
$ I = e^{x} \tan^{-1} x + c $.

(A) Let $I = \int e^{\sin x} \cdot \left(\frac{\sin x+1}{\sec x}\right) d x$.
Since $\frac{1}{\sec x} = \cos x$,we can rewrite the integral as:
$I = \int e^{\sin x} (\sin x + 1) \cos x \, dx$.
Let $u = \sin x$,then $du = \cos x \, dx$.
Substituting these into the integral:
$I = \int e^u (u + 1) \, du$.
$I = \int (u e^u + e^u) \, du$.
Using the integration by parts formula $\int (f(u) + f'(u)) e^u \, du = e^u f(u) + C$,where $f(u) = u$ and $f'(u) = 1$:
$I = e^u \cdot u + C$.
Substituting $u = \sin x$ back:
$I = \sin x \cdot e^{\sin x} + C$.

(B) Let $I = \int \frac{x-1}{(x+1)^{3}} e^{x} d x$.
We can rewrite the numerator as $(x+1)-2$:
$I = \int \frac{(x+1)-2}{(x+1)^{3}} e^{x} d x$
$I = \int \left( \frac{x+1}{(x+1)^{3}} - \frac{2}{(x+1)^{3}} \right) e^{x} d x$
$I = \int \frac{e^{x}}{(x+1)^{2}} d x - \int \frac{2 e^{x}}{(x+1)^{3}} d x$.
Now,apply integration by parts to the first integral $\int \frac{e^{x}}{(x+1)^{2}} d x$,taking $u = \frac{1}{(x+1)^{2}}$ and $dv = e^{x} dx$:
$du = -2(x+1)^{-3} dx$ and $v = e^{x}$.
$\int \frac{e^{x}}{(x+1)^{2}} d x = \frac{e^{x}}{(x+1)^{2}} - \int e^{x} \left( -\frac{2}{(x+1)^{3}} \right) d x$
$= \frac{e^{x}}{(x+1)^{2}} + \int \frac{2 e^{x}}{(x+1)^{3}} d x$.
Substituting this back into the expression for $I$:
$I = \left( \frac{e^{x}}{(x+1)^{2}} + \int \frac{2 e^{x}}{(x+1)^{3}} d x \right) - \int \frac{2 e^{x}}{(x+1)^{3}} d x$
$I = \frac{e^{x}}{(x+1)^{2}} + C$.

(C) We are given the integral $I = \int e^{x} \left( \frac{\sin x + \cos x}{\cos^2 x} \right) dx$.
Using the identity $1 - \sin^2 x = \cos^2 x$,we rewrite the expression:
$I = \int e^{x} \left( \frac{\sin x}{\cos^2 x} + \frac{\cos x}{\cos^2 x} \right) dx$
$I = \int e^{x} (\tan x \sec x + \sec x) dx$
Let $f(x) = \sec x$. Then $f'(x) = \sec x \tan x$.
We know the standard integral formula $\int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + C$.
Applying this to our integral:
$I = \int e^{x} (\sec x + \sec x \tan x) dx = e^{x} \sec x + C$.

(A) Let $I = \int e^{\tan ^{-1} x} \left(1 + \frac{x}{1+x^2}\right) dx$.
We can split the integral as $I = \int e^{\tan ^{-1} x} dx + \int \frac{x e^{\tan ^{-1} x}}{1+x^2} dx$.
Now,apply integration by parts to the first integral $\int e^{\tan ^{-1} x} dx$ by taking $u = e^{\tan ^{-1} x}$ and $dv = dx$.
Then $du = e^{\tan ^{-1} x} \cdot \frac{1}{1+x^2} dx$ and $v = x$.
Using the formula $\int u dv = uv - \int v du$,we get:
$\int e^{\tan ^{-1} x} dx = x e^{\tan ^{-1} x} - \int \frac{x e^{\tan ^{-1} x}}{1+x^2} dx$.
Substituting this back into the expression for $I$:
$I = \left(x e^{\tan ^{-1} x} - \int \frac{x e^{\tan ^{-1} x}}{1+x^2} dx\right) + \int \frac{x e^{\tan ^{-1} x}}{1+x^2} dx + c$.
The integral terms cancel out,leaving $I = x e^{\tan ^{-1} x} + c$.

(B) We know that $\int e^{x}(f(x)+f'(x))dx = e^{x}f(x)+c$.
Here,let $f(x) = x^{5}$.
Then,$f'(x) = 5x^{4}$.
The given integral is $\int e^{x}(x^{5}+5x^{4}+1)dx$.
We can rewrite this as $\int e^{x}(x^{5}+5x^{4})dx + \int e^{x}dx$.
Using the formula $\int e^{x}(f(x)+f'(x))dx = e^{x}f(x)+c$,we get $\int e^{x}(x^{5}+5x^{4})dx = e^{x}x^{5}+c_1$.
Thus,the total integral is $e^{x}x^{5} + e^{x} + c$.

(D) To evaluate $I = \int_0^1 \frac{x e^x}{(2+x)^3} d x$,we rewrite the integrand as follows:
$\frac{x}{(2+x)^3} = \frac{(x+2)-2}{(2+x)^3} = \frac{1}{(2+x)^2} - \frac{2}{(2+x)^3}$.
Let $f(x) = \frac{1}{(2+x)^2}$. Then $f'(x) = -2(2+x)^{-3} = -\frac{2}{(2+x)^3}$.
Thus,the integral becomes $\int_0^1 e^x [f(x) + f'(x)] d x$.
Using the standard result $\int e^x [f(x) + f'(x)] d x = e^x f(x) + C$,we get:
$I = [e^x \cdot \frac{1}{(2+x)^2}]_0^1$.
Evaluating at the limits:
$I = (e^1 \cdot \frac{1}{(2+1)^2}) - (e^0 \cdot \frac{1}{(2+0)^2}) = \frac{e}{9} - \frac{1}{4}$.

(C) We are given the integral $ I = \int \frac{(x+3) e^{x}}{(x+4)^{2}} d x $.
To solve this,we rewrite the numerator as $(x+4-1)$:
$ I = \int e^{x} \frac{(x+4-1)}{(x+4)^{2}} d x $.
Splitting the fraction,we get:
$ I = \int e^{x} \left[ \frac{x+4}{(x+4)^{2}} - \frac{1}{(x+4)^{2}} \right] d x $.
$ I = \int e^{x} \left[ \frac{1}{x+4} - \frac{1}{(x+4)^{2}} \right] d x $.
We know the standard integration formula:
$ \int e^{x} (f(x) + f'(x)) d x = e^{x} f(x) + C $.
Here,let $ f(x) = \frac{1}{x+4} $.
Then,$ f'(x) = -\frac{1}{(x+4)^{2}} $.
Substituting these into the formula,we get:
$ I = e^{x} \left( \frac{1}{x+4} \right) + C = \frac{e^{x}}{x+4} + C $.

(C) We know the standard integral formula: $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$.
Here,let $f(x) = \log x$.
Then,$f'(x) = \frac{1}{x}$.
Substituting these into the formula:
$\int e^x \left( \log x + \frac{1}{x} \right) dx = e^x \log x + C$.
However,looking at the provided options,the question likely intended to be $\int e^x \left( \log x + \frac{1}{x} \right) dx$. If the question is $\int e^x \left( \log x + \frac{1}{x} \right) dx$,the result is $e^x \log x + C$. If the question is $\int e^x \left( \frac{1}{x} - \frac{1}{x^2} \right) dx$,the result is $e^x \frac{1}{x} + C$. Given the structure of the options,there is a mismatch. Assuming the standard form $\int e^x (f(x) + f'(x)) dx$,the closest match for the form $e^x(\log x + \dots)$ is option $C$ if the derivative was different.

(A) We have the integral $I = \int e^{-2x} \left( \frac{1 - \sin 2x}{1 + \cos 2x} \right) dx$.
Using trigonometric identities,$1 - \sin 2x = 1 - 2 \sin x \cos x$ and $1 + \cos 2x = 2 \cos^2 x$.
So,the expression becomes $\frac{1 - 2 \sin x \cos x}{2 \cos^2 x} = \frac{1}{2} \sec^2 x - \tan x$.
Thus,$I = \int e^{-2x} (\frac{1}{2} \sec^2 x - \tan x) dx = \frac{1}{2} \int e^{-2x} \sec^2 x dx - \int e^{-2x} \tan x dx$.
Applying integration by parts to the first term: $\int e^{-2x} \sec^2 x dx = e^{-2x} \tan x - \int (-2e^{-2x}) \tan x dx = e^{-2x} \tan x + 2 \int e^{-2x} \tan x dx$.
Substituting this back into $I$: $I = \frac{1}{2} [e^{-2x} \tan x + 2 \int e^{-2x} \tan x dx] - \int e^{-2x} \tan x dx$.
$I = \frac{1}{2} e^{-2x} \tan x + \int e^{-2x} \tan x dx - \int e^{-2x} \tan x dx = \frac{1}{2} e^{-2x} \tan x + C$.

(C) Let $I = \int \left( \frac{\log x - 1}{1 + (\log x)^2} \right)^2 dx$.
Substitute $t = \log x$,then $x = e^t$ and $dx = e^t dt$.
$I = \int e^t \frac{(t - 1)^2}{(1 + t^2)^2} dt = \int e^t \frac{t^2 - 2t + 1}{(1 + t^2)^2} dt$.
$I = \int e^t \left( \frac{t^2 + 1 - 2t}{(1 + t^2)^2} \right) dt = \int e^t \left( \frac{1}{1 + t^2} - \frac{2t}{(1 + t^2)^2} \right) dt$.
Using the formula $\int e^t [f(t) + f'(t)] dt = e^t f(t) + C$,where $f(t) = \frac{1}{1 + t^2}$ and $f'(t) = -\frac{2t}{(1 + t^2)^2}$.
Thus,$I = e^t \left( \frac{1}{1 + t^2} \right) + C = \frac{x}{1 + (\log x)^2} + C$.

(D) Let $I = \int \frac{e^{\sin x}(\sin 2x - 8 \cos x)}{2(\sin x - 3)^2} dx$.
Using $\sin 2x = 2 \sin x \cos x$,we have:
$I = \int \frac{e^{\sin x}(2 \sin x \cos x - 8 \cos x)}{2(\sin x - 3)^2} dx = \int \frac{e^{\sin x} \cos x (\sin x - 4)}{(\sin x - 3)^2} dx$.
Let $t = \sin x$,then $dt = \cos x dx$.
$I = \int \frac{e^t (t - 4)}{(t - 3)^2} dt = \int e^t \left( \frac{t - 3 - 1}{(t - 3)^2} \right) dt = \int e^t \left( \frac{1}{t - 3} - \frac{1}{(t - 3)^2} \right) dt$.
Using the formula $\int e^t (f(t) + f'(t)) dt = e^t f(t) + c$,where $f(t) = \frac{1}{t - 3}$ and $f'(t) = -\frac{1}{(t - 3)^2}$.
Thus,$I = e^t \left( \frac{1}{t - 3} \right) + c = \frac{e^{\sin x}}{\sin x - 3} + c$.

(C) Given the integral $\int e^{\sin x}(1+\sec x \tan x) d x = e^{\sin x} f(x) + c$.
Let $I = \int e^{\sin x}(1+\sec x \tan x) d x$.
We can rewrite the integrand as $e^{\sin x} + e^{\sin x} \sec x \tan x$.
This does not immediately fit the form $\int e^{g(x)}(g'(x) + h(x)) dx$.
Let us differentiate $e^{\sin x} f(x)$ with respect to $x$:
$\frac{d}{dx} [e^{\sin x} f(x)] = e^{\sin x} \cos x f(x) + e^{\sin x} f'(x) = e^{\sin x} (f'(x) + f(x) \cos x)$.
Comparing this with the integrand $e^{\sin x}(1+\sec x \tan x)$,we have $f'(x) + f(x) \cos x = 1 + \sec x \tan x$.
By inspection,if $f(x) = \sec x$,then $f'(x) = \sec x \tan x$.
Substituting this into the equation: $\sec x \tan x + \sec x \cos x = \sec x \tan x + 1$.
This matches the integrand. Thus,$f(x) = \sec x$.
We need to find the number of solutions for $f(x) = 1$ in $0 \leq x \leq 2 \pi$.
$\sec x = 1 \implies \cos x = 1$.
In the interval $[0, 2 \pi]$,$\cos x = 1$ at $x = 0$ and $x = 2 \pi$.
Thus,there are $2$ solutions.

(B) Let $I = \int \frac{e^{\cot x}}{\sin^2 x} (2 \log \csc x + \sin 2 x) dx$.
Substitute $t = \cot x$,then $dt = -\csc^2 x dx$,which implies $\csc^2 x dx = -dt$.
Also,$\sin 2x = 2 \sin x \cos x = \frac{2 \sin x \cos x}{\sin^2 x} \cdot \sin^2 x = 2 \cot x \sin^2 x$.
Wait,let us simplify the integrand: $\frac{1}{\sin^2 x} = \csc^2 x$.
So,$I = \int e^{\cot x} \csc^2 x (2 \log \csc x + 2 \sin x \cos x) dx$.
Since $\sin 2x = 2 \sin x \cos x$,we have $\frac{\sin 2x}{\sin^2 x} = 2 \cot x$.
Thus,$I = \int e^{\cot x} (2 \csc^2 x \log \csc x + 2 \cot x \csc^2 x) dx$.
Let $u = \cot x$,then $du = -\csc^2 x dx$.
Note that $\csc^2 x = 1 + \cot^2 x = 1 + u^2$.
Also $\log \csc x = \log (1 + u^2)^{1/2} = \frac{1}{2} \log (1 + u^2)$.
Substituting these into the integral:
$I = -\int e^u (2 \cdot \frac{1}{2} \log (1 + u^2) + 2u) du = -\int e^u (\log (1 + u^2) + 2u) du$.
This does not simplify directly to the standard form $\int e^u (f(u) + f'(u)) du$.
Let us re-evaluate: $\frac{d}{dx} (e^{\cot x} \log \csc^2 x) = e^{\cot x} (-\csc^2 x) \log \csc^2 x + e^{\cot x} \frac{1}{\csc^2 x} (2 \csc x \cdot -\csc x \cot x) = -e^{\cot x} \csc^2 x \log \csc^2 x - 2 e^{\cot x} \cot x$.
Comparing with the integrand,the correct form is $-2 e^{\cot x} \log \csc x + C$.

(B) Let $I = \int e^{x \operatorname{cosec} x} \cdot \operatorname{cosec} x \cdot (1 - x \cot x) \, dx$.
Substitute $t = x \operatorname{cosec} x$.
Differentiating with respect to $x$ using the product rule:
$\frac{dt}{dx} = x \cdot \frac{d}{dx}(\operatorname{cosec} x) + \operatorname{cosec} x \cdot \frac{d}{dx}(x)$
$\frac{dt}{dx} = x(-\operatorname{cosec} x \cot x) + \operatorname{cosec} x(1)$
$\frac{dt}{dx} = \operatorname{cosec} x(1 - x \cot x)$
Therefore,$dt = \operatorname{cosec} x(1 - x \cot x) \, dx$.
Substituting these into the integral:
$I = \int e^t \, dt = e^t + c$
Substituting back $t = x \operatorname{cosec} x$:
$I = e^{x \operatorname{cosec} x} + c$.
Thus,option $B$ is correct.

(C) Let $I = \int \frac{\log _e x}{(1+\log _e x)^2} dx$.
Substitute $\log _e x = t$,which implies $x = e^t$ and $dx = e^t dt$.
Substituting these into the integral,we get:
$I = \int \frac{t}{(1+t)^2} e^t dt$.
We can rewrite the numerator as $(t+1-1)$:
$I = \int \frac{t+1-1}{(1+t)^2} e^t dt = \int \left( \frac{1}{1+t} - \frac{1}{(1+t)^2} \right) e^t dt$.
Using the standard integral formula $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$,where $f(t) = \frac{1}{1+t}$ and $f'(t) = -\frac{1}{(1+t)^2}$:
$I = e^t \left( \frac{1}{1+t} \right) + C$.
Substituting $t = \log _e x$ back into the equation:
$I = e^{\log _e x} \left( \frac{1}{1+\log _e x} \right) + C = \frac{x}{1+\log _e x} + C$.

(C) We have $I = \int e^x \left(\frac{x+2}{x+4}\right)^2 dx$.
Rewrite the numerator as $(x+4-2)$:
$I = \int e^x \left(\frac{x+4-2}{x+4}\right)^2 dx = \int e^x \left(1 - \frac{2}{x+4}\right)^2 dx$.
Expanding the square:
$I = \int e^x \left(1 - \frac{4}{x+4} + \frac{4}{(x+4)^2}\right) dx$.
This can be written as:
$I = \int e^x \left(1 - \frac{4}{x+4}\right) dx + \int \frac{4 e^x}{(x+4)^2} dx$.
Let $f(x) = 1 - \frac{4}{x+4}$. Then $f'(x) = -(-4)(x+4)^{-2} = \frac{4}{(x+4)^2}$.
Using the standard integral formula $\int e^x (f(x) + f'(x)) dx = e^x f(x) + c$:
$I = e^x \left(1 - \frac{4}{x+4}\right) + c = e^x \left(\frac{x+4-4}{x+4}\right) + c = \frac{x e^x}{x+4} + c$.

(A) We use the formula $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c$.
Let $f(x) = (x+1)^2 = x^2 + 2x + 1$.
Then $f'(x) = 2x + 2 = 2(x+1)$.
This does not directly fit the form.
Alternatively,expand the expression:
$\int e^x(x^2+2x+1) dx = \int e^x x^2 dx + \int e^x(2x+1) dx$.
Using integration by parts on $\int e^x x^2 dx$:
$= x^2 e^x - \int 2x e^x dx + \int 2x e^x dx + \int e^x dx = x^2 e^x + e^x + c = e^x(x^2+1) + c$.

(A) Let $I = \int \frac{e^{\tan ^{-1} x}}{1+x^2} \left[ (\sec ^{-1} \sqrt{1+x^2})^2 + \cos ^{-1} \left( \frac{1-x^2}{1+x^2} \right) \right] dx$.
Since $\sec ^{-1} \sqrt{1+x^2} = \tan ^{-1} x$ and $\cos ^{-1} \left( \frac{1-x^2}{1+x^2} \right) = 2 \tan ^{-1} x$,the integral becomes:
$I = \int \frac{e^{\tan ^{-1} x}}{1+x^2} [(\tan ^{-1} x)^2 + 2 \tan ^{-1} x] dx$.
Let $t = \tan ^{-1} x$,then $dt = \frac{1}{1+x^2} dx$.
$I = \int e^t (t^2 + 2t) dt = \int (t^2 e^t + 2t e^t) dt$.
Using the formula $\int e^t (f(t) + f'(t)) dt = e^t f(t) + C$,where $f(t) = t^2$ and $f'(t) = 2t$:
$I = e^t t^2 + C = e^{\tan ^{-1} x} (\tan ^{-1} x)^2 + C$.

(A) We are given the integral $I = \int \frac{e^x(x + 3)}{(x + 5)^3} dx$.
First,we rewrite the numerator $(x + 3)$ as $(x + 5 - 2)$:
$I = \int \frac{e^x(x + 5 - 2)}{(x + 5)^3} dx$
$I = \int e^x \left[ \frac{x + 5}{(x + 5)^3} - \frac{2}{(x + 5)^3} \right] dx$
$I = \int e^x \left[ \frac{1}{(x + 5)^2} - \frac{2}{(x + 5)^3} \right] dx$
We use the standard integration formula $\int e^x (f(x) + f'(x)) dx = e^x f(x) + c$.
Let $f(x) = \frac{1}{(x + 5)^2} = (x + 5)^{-2}$.
Then $f'(x) = -2(x + 5)^{-3} = -\frac{2}{(x + 5)^3}$.
Since the integrand is in the form $e^x(f(x) + f'(x))$,the integral is $e^x f(x) + c$.
Thus,$I = \frac{e^x}{(x + 5)^2} + c$.

(A) Let $I = \int \frac{e^{\tan ^{-1} x}}{1+x^2} \left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2 + \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] d x$ for $x > 0$.
Substitute $\tan ^{-1} x = \theta$,so $x = \tan \theta$ and $\frac{1}{1+x^2} d x = d \theta$.
Since $x > 0$,$\theta \in (0, \pi/2)$.
Note that $\sec ^{-1} \sqrt{1+x^2} = \sec ^{-1} \sqrt{1+\tan^2 \theta} = \sec ^{-1} \sec \theta = \theta$.
Also,$\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) = \cos ^{-1}(\cos 2\theta) = 2\theta$ (since $2\theta \in (0, \pi)$).
Substituting these into the integral:
$I = \int e^{\theta} [\theta^2 + 2\theta] d \theta$.
Using the standard integral formula $\int e^x [f(x) + f'(x)] d x = e^x f(x) + c$,where $f(\theta) = \theta^2$ and $f'(\theta) = 2\theta$:
$I = e^{\theta} \theta^2 + c$.
Substituting back $\theta = \tan ^{-1} x$:
$I = e^{\tan ^{-1} x} (\tan ^{-1} x)^2 + c$.

(C) Let $I = \int 3^x \left(f^{\prime}(x) + f(x) \log 3\right) dx$.
We can rewrite the integrand as:
$I = \int \left(3^x f^{\prime}(x) + 3^x f(x) \log 3\right) dx$.
Recall the product rule for differentiation: $\frac{d}{dx} \left(3^x f(x)\right) = 3^x f^{\prime}(x) + f(x) \cdot \frac{d}{dx}(3^x) = 3^x f^{\prime}(x) + f(x) \cdot 3^x \log 3$.
Thus,the integrand is the derivative of $3^x f(x)$.
Therefore,$\int \frac{d}{dx} \left(3^x f(x)\right) dx = 3^x f(x) + c$.

(A) We are given the integral $I = \int e^x \left(\frac{x+2}{x+4}\right)^2 dx$.
First,rewrite the expression inside the square:
$\frac{x+2}{x+4} = \frac{x+4-2}{x+4} = 1 - \frac{2}{x+4}$.
Thus,$\left(\frac{x+2}{x+4}\right)^2 = \left(1 - \frac{2}{x+4}\right)^2 = 1 - \frac{4}{x+4} + \frac{4}{(x+4)^2}$.
Now,the integral becomes:
$I = \int e^x \left(1 - \frac{4}{x+4} + \frac{4}{(x+4)^2}\right) dx$.
We know the standard form $\int e^x [g(x) + g'(x)] dx = e^x g(x) + C$.
Let $g(x) = 1 - \frac{4}{x+4}$.
Then $g'(x) = -\left(-\frac{4}{(x+4)^2}\right) = \frac{4}{(x+4)^2}$.
Therefore,$I = e^x \left(1 - \frac{4}{x+4}\right) + C = e^x \left(\frac{x+4-4}{x+4}\right) + C = \frac{x e^x}{x+4} + C$.
Hence,$f(x) = \frac{x e^x}{x+4}$.

(A) Let $I = \int \left( \frac{2 - \sin 2x}{1 - \cos 2x} \right) e^x \, dx$.
Using trigonometric identities $\sin 2x = 2 \sin x \cos x$ and $1 - \cos 2x = 2 \sin^2 x$,we get:
$I = \int \left( \frac{2 - 2 \sin x \cos x}{2 \sin^2 x} \right) e^x \, dx$
$I = \int \left( \frac{1}{\sin^2 x} - \frac{\sin x \cos x}{\sin^2 x} \right) e^x \, dx$
$I = \int (\operatorname{cosec}^2 x - \cot x) e^x \, dx$
$I = \int e^x \operatorname{cosec}^2 x \, dx - \int e^x \cot x \, dx$.
Applying integration by parts to the first integral $\int e^x \operatorname{cosec}^2 x \, dx$:
Let $u = \cot x$,then $du = -\operatorname{cosec}^2 x \, dx$.
$\int e^x \operatorname{cosec}^2 x \, dx = e^x(-\cot x) - \int e^x(-\cot x) \, dx = -e^x \cot x + \int e^x \cot x \, dx$.
Substituting this back into the expression for $I$:
$I = (-e^x \cot x + \int e^x \cot x \, dx) - \int e^x \cot x \, dx + c$
$I = -e^x \cot x + c$.

(C) Let $I = \int e^x \frac{x^2+1}{(x+1)^2} d x$.
We can rewrite the numerator as $x^2 - 1 + 2$.
$I = \int e^x \left( \frac{x^2-1+2}{(x+1)^2} \right) d x = \int e^x \left( \frac{(x-1)(x+1)}{(x+1)^2} + \frac{2}{(x+1)^2} \right) d x$.
$I = \int e^x \left( \frac{x-1}{x+1} + \frac{2}{(x+1)^2} \right) d x$.
Let $f(x) = \frac{x-1}{x+1}$.
Then $f'(x) = \frac{(x+1)(1) - (x-1)(1)}{(x+1)^2} = \frac{x+1-x+1}{(x+1)^2} = \frac{2}{(x+1)^2}$.
Using the standard integral formula $\int e^x \{f(x) + f'(x)\} d x = e^x f(x) + C$,we get:
$I = e^x \left( \frac{x-1}{x+1} \right) + C$.

(A) Given the integral $I = \int \frac{3-x^2}{1-2 x+x^2} e^x d x$.
Since $1-2x+x^2 = (1-x)^2$,we can rewrite the integrand as:
$I = \int \frac{3-x^2}{(1-x)^2} e^x d x$
We can manipulate the numerator to match the form $f(x) + f'(x)$:
$I = \int \frac{2 + 1 - x^2}{(1-x)^2} e^x d x = \int \left( \frac{2}{(1-x)^2} + \frac{1-x^2}{(1-x)^2} \right) e^x d x$
$I = \int \left( \frac{2}{(1-x)^2} + \frac{(1-x)(1+x)}{(1-x)^2} \right) e^x d x$
$I = \int \left( \frac{2}{(1-x)^2} + \frac{1+x}{1-x} \right) e^x d x$
Let $f(x) = \frac{1+x}{1-x}$. Then $f'(x) = \frac{(1-x)(1) - (1+x)(-1)}{(1-x)^2} = \frac{1-x+1+x}{(1-x)^2} = \frac{2}{(1-x)^2}$.
Since $\int (f(x) + f'(x)) e^x d x = e^x f(x) + c$,we have:
$I = e^x \left( \frac{1+x}{1-x} \right) + c$.
Comparing this with the given expression $e^x f(x) + c$,we get $f(x) = \frac{1+x}{1-x}$.

(B) We have the integral $I = \int e^{x / 2} \left( \frac{2 + \sin x}{1 + \cos x} \right) dx$.
Using the half-angle formulas $\sin x = \frac{2 \tan(x/2)}{1 + \tan^2(x/2)}$ and $1 + \cos x = 2 \cos^2(x/2)$,we can simplify the integrand:
$\frac{2 + \sin x}{1 + \cos x} = \frac{2}{2 \cos^2(x/2)} + \frac{2 \tan(x/2)}{(1 + \tan^2(x/2)) \cdot 2 \cos^2(x/2)}$
$= \sec^2(x/2) + \frac{\tan(x/2)}{\sec^2(x/2) \cdot \cos^2(x/2)} = \sec^2(x/2) + \tan(x/2)$.
Thus,$I = \int e^{x/2} (\sec^2(x/2) + \tan(x/2)) dx$.
Let $f(x) = \tan(x/2)$,then $f'(x) = \sec^2(x/2) \cdot \frac{1}{2}$.
This does not directly fit the form $\int e^x (f(x) + f'(x)) dx$.
Let $u = x/2$,then $du = dx/2$,so $dx = 2du$.
$I = \int e^u (\sec^2 u + \tan u) \cdot 2 du = 2 \int e^u (\tan u + \sec^2 u) du$.
Since $\frac{d}{du}(\tan u) = \sec^2 u$,using the formula $\int e^u (f(u) + f'(u)) du = e^u f(u) + c$,we get:
$I = 2 e^u \tan u + c = 2 e^{x/2} \tan(x/2) + c$.
Hence,option $B$ is correct.

(D) Let $I = \int \left( \frac{1}{\log x} - \frac{1}{(\log x)^2} \right) dx$.
Substitute $\log x = t$,which implies $x = e^t$.
Then,$dx = e^t dt$.
Substituting these into the integral,we get:
$I = \int \left( \frac{1}{t} - \frac{1}{t^2} \right) e^t dt$.
We know the standard integration formula: $\int e^t (f(t) + f'(t)) dt = e^t f(t) + C$.
Here,$f(t) = \frac{1}{t}$ and $f'(t) = -\frac{1}{t^2}$.
Therefore,$I = e^t \cdot \frac{1}{t} + c$.
Substituting back $t = \log x$ and $e^t = x$:
$I = \frac{x}{\log x} + c$.
Thus,option $D$ is correct.

(A) Assertion: Let $I = \int_2^e \left(\frac{1}{\log_e x} - \frac{1}{(\log_e x)^2}\right) dx$.
Let $\log_e x = y$,then $x = e^y$ and $dx = e^y dy$.
When $x = 2$,$y = \log_e 2$. When $x = e$,$y = 1$.
Substituting these into the integral: $I = \int_{\log_e 2}^1 e^y \left(\frac{1}{y} - \frac{1}{y^2}\right) dy$.
Using the formula $\int e^y (f(y) + f'(y)) dy = e^y f(y) + C$,where $f(y) = \frac{1}{y}$ and $f'(y) = -\frac{1}{y^2}$.
$I = \left[ e^y \cdot \frac{1}{y} \right]_{\log_e 2}^1 = \left( e^1 \cdot \frac{1}{1} \right) - \left( e^{\log_e 2} \cdot \frac{1}{\log_e 2} \right) = e - \frac{2}{\log_e 2} = e - 2 \log_2 e$.
The Reason $(R)$ provides the standard formula $\int_a^b e^x (f(x) + f'(x)) dx = [e^x f(x)]_a^b$,which is correct.
Thus,both Assertion and Reason are true,and the Reason is the correct explanation of the Assertion.

(B) $I = \int_0^1 \frac{x e^x}{(x+1)^2} d x$
We can rewrite the integrand as:
$I = \int_0^1 e^x \left[ \frac{x+1-1}{(x+1)^2} \right] d x$
$I = \int_0^1 e^x \left[ \frac{1}{x+1} - \frac{1}{(x+1)^2} \right] d x$
Using the standard integral formula $\int e^x [f(x) + f'(x)] d x = e^x f(x) + C$,let $f(x) = \frac{1}{x+1}$.
Then,$f'(x) = -\frac{1}{(x+1)^2}$.
Thus,the integral becomes:
$I = \left[ e^x \cdot \frac{1}{x+1} \right]_0^1$
$I = \left( \frac{e^1}{1+1} \right) - \left( \frac{e^0}{0+1} \right)$
$I = \frac{e}{2} - 1$

(B) We have the integral $I = \int e^x \left( \frac{2 + \sin 2x}{1 + \cos 2x} \right) dx$.
Using the trigonometric identities $\sin 2x = 2 \sin x \cos x$ and $1 + \cos 2x = 2 \cos^2 x$,we get:
$I = \int e^x \left( \frac{2 + 2 \sin x \cos x}{2 \cos^2 x} \right) dx$
$I = \int e^x \left( \frac{2(1 + \sin x \cos x)}{2 \cos^2 x} \right) dx$
$I = \int e^x \left( \frac{1}{\cos^2 x} + \frac{\sin x \cos x}{\cos^2 x} \right) dx$
$I = \int e^x (\sec^2 x + \tan x) dx$.
Recall the standard integral form $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$.
Here,let $f(x) = \tan x$,then $f'(x) = \sec^2 x$.
Therefore,$I = e^x \tan x + C$.

(D) Let $I = \int_{\alpha+1}^{\alpha} \frac{e^x(\alpha-x)}{(x-\alpha+1)^2} dx$.
Substitute $u = x - \alpha + 1$,then $du = dx$.
When $x = \alpha+1$,$u = 2$. When $x = \alpha$,$u = 1$.
Also,$\alpha - x = 1 - u$.
So,$I = \int_{2}^{1} \frac{e^{u+\alpha-1}(1-u)}{u^2} du = e^{\alpha-1} \int_{2}^{1} e^u \left(\frac{1}{u^2} - \frac{1}{u}\right) du$.
Using the formula $\int e^u (f(u) + f'(u)) du = e^u f(u) + C$,where $f(u) = -\frac{1}{u}$ and $f'(u) = \frac{1}{u^2}$.
$I = e^{\alpha-1} \left[ e^u \left(-\frac{1}{u}\right) \right]_{2}^{1} = e^{\alpha-1} \left[ -e^1 + \frac{e^2}{2} \right] = e^{\alpha-1} \left[ \frac{e^2 - 2e}{2} \right] = e^{\alpha} \left( \frac{e-2}{2} \right)$.

(B) Given,$\int e^{\alpha x}\left(\frac{1-\beta \sin x}{1-\cos x}\right) d x=-e^x \cot \frac{x}{2}+c$.
By differentiating both sides with respect to $x$,we get:
$e^{\alpha x}\left(\frac{1-\beta \sin x}{1-\cos x}\right) = \frac{d}{dx} \left(-e^x \cot \frac{x}{2}\right) + 0$.
Using the product rule,$\frac{d}{dx}(uv) = u'v + uv'$:
$e^{\alpha x}\left(\frac{1-\beta \sin x}{1-\cos x}\right) = -\left[e^x \cot \frac{x}{2} + e^x \left(-\frac{1}{2} \csc^2 \frac{x}{2}\right)\right]$.
$e^{\alpha x}\left(\frac{1-\beta \sin x}{1-\cos x}\right) = -e^x \left[\frac{\cos(x/2)}{\sin(x/2)} - \frac{1}{2 \sin^2(x/2)}\right]$.
$e^{\alpha x}\left(\frac{1-\beta \sin x}{1-\cos x}\right) = -e^x \left[\frac{2 \sin(x/2) \cos(x/2) - 1}{2 \sin^2(x/2)}\right]$.
Using $2 \sin(x/2) \cos(x/2) = \sin x$ and $2 \sin^2(x/2) = 1 - \cos x$:
$e^{\alpha x}\left(\frac{1-\beta \sin x}{1-\cos x}\right) = -e^x \left[\frac{\sin x - 1}{1 - \cos x}\right] = e^x \left(\frac{1 - \sin x}{1 - \cos x}\right)$.
Comparing both sides,we get $\alpha = 1$ and $\beta = 1$.
Therefore,$\frac{\alpha^2 + \beta^2}{2 \alpha \beta} = \frac{1^2 + 1^2}{2(1)(1)} = \frac{2}{2} = 1$.

(D) Let $I = \int e^x \left( \frac{\sec^2 x + \tan x - \cot x}{\sin x} \right) dx$.
We can rewrite the integrand as:
$I = \int e^x \left( \frac{\sec^2 x}{\sin x} + \frac{\tan x}{\sin x} - \frac{\cot x}{\sin x} \right) dx$
$I = \int e^x \left( \sec^2 x \operatorname{cosec} x + \sec x - \cot x \operatorname{cosec} x \right) dx$
Using the identity $\sec^2 x = 1 + \tan^2 x$,we have:
$I = \int e^x \left( \operatorname{cosec} x(1 + \tan^2 x) + \sec x - \cot x \operatorname{cosec} x \right) dx$
$I = \int e^x \left( \operatorname{cosec} x + \operatorname{cosec} x \tan^2 x + \sec x - \cot x \operatorname{cosec} x \right) dx$
Since $\operatorname{cosec} x \tan^2 x = \frac{1}{\sin x} \cdot \frac{\sin^2 x}{\cos^2 x} = \frac{\sin x}{\cos^2 x} = \sec x \tan x$,the expression becomes:
$I = \int e^x \left( (\operatorname{cosec} x - \cot x \operatorname{cosec} x) + (\sec x + \sec x \tan x) \right) dx$
This is of the form $\int e^x (f(x) + f'(x)) dx = e^x f(x) + c$.
Here,$f(x) = \operatorname{cosec} x + \sec x$,and $f'(x) = -\operatorname{cosec} x \cot x + \sec x \tan x$.
Thus,$I = e^x(\operatorname{cosec} x + \sec x) + c$.
Therefore,option $(D)$ is correct.

(C) Let $I = \int \left( \frac{1-\log x}{1+(\log x)^2} \right)^2 dx$.
Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$.
Then,$I = \int \left( \frac{1-t}{1+t^2} \right)^2 e^t dt$.
This is of the form $\int e^t [f(t) + f'(t)] dt = e^t f(t) + c$.
Let $f(t) = \frac{t}{1+t^2}$. Then $f'(t) = \frac{(1+t^2)(1) - t(2t)}{(1+t^2)^2} = \frac{1-t^2}{(1+t^2)^2}$.
This does not match directly. Let us rewrite the integral:
$I = \int e^t \frac{(1-t)^2}{(1+t^2)^2} dt = \int e^t \frac{1-2t+t^2}{(1+t^2)^2} dt = \int e^t \left( \frac{1+t^2-2t}{(1+t^2)^2} \right) dt = \int e^t \left( \frac{1}{1+t^2} - \frac{2t}{(1+t^2)^2} \right) dt$.
Here,if $f(t) = \frac{1}{1+t^2}$,then $f'(t) = \frac{-2t}{(1+t^2)^2}$.
Thus,$I = e^t \left( \frac{1}{1+t^2} \right) + c$.
Substituting back $t = \log x$,we get $I = \frac{e^{\log x}}{1+(\log x)^2} + c = \frac{x}{1+(\log x)^2} + c$.

(B) We use the standard integral formula: $\int e^{ax}(\sin bx - \cos bx) dx = \frac{e^{ax}}{a^2+b^2} [a \sin bx - b \cos bx - (a \cos bx + b \sin bx)] + c = \frac{e^{ax}}{a^2+b^2} [(a-b) \sin bx - (a+b) \cos bx] + c$.
Here,$a = 4$ and $b = 3$.
Substituting these values:
$\int e^{4x}(\sin 3x - \cos 3x) dx = \frac{e^{4x}}{4^2+3^2} [(4-3) \sin 3x - (4+3) \cos 3x] + c$
$= \frac{e^{4x}}{16+9} [1 \sin 3x - 7 \cos 3x] + c$
$= \frac{e^{4x}}{25}(\sin 3x - 7 \cos 3x) + c$.
Thus,the correct option is $B$.

(A) We use the formula for the integral of the form $\int e^{ax} f(x) dx = \frac{e^{ax}}{a} [f(x) - \frac{f'(x)}{a} + \frac{f''(x)}{a^2} - \frac{f'''(x)}{a^3} + \dots]$.
Here,$a = -1$ and $f(x) = x^3 - 2x^2 + 3x - 4$.
Calculating the derivatives:
$f'(x) = 3x^2 - 4x + 3$
$f''(x) = 6x - 4$
$f'''(x) = 6$
$f^{(4)}(x) = 0$
Substituting these into the formula:
$\int e^{-x} f(x) dx = \frac{e^{-x}}{-1} [f(x) - \frac{f'(x)}{-1} + \frac{f''(x)}{(-1)^2} - \frac{f'''(x)}{(-1)^3}] + c$
$= -e^{-x} [f(x) + f'(x) + f''(x) + f'''(x)] + c$
$= -e^{-x} [(x^3 - 2x^2 + 3x - 4) + (3x^2 - 4x + 3) + (6x - 4) + 6] + c$
$= -e^{-x} [x^3 + (-2+3)x^2 + (3-4+6)x + (-4+3-4+6)] + c$
$= -e^{-x} [x^3 + x^2 + 5x + 1] + c$.