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

(N/A) Let $I = \int e^{x}(\sin x + \cos x) \, dx$.
We know the standard integral formula: $\int e^{x} \{f(x) + f'(x)\} \, dx = e^{x} f(x) + C$.
Let $f(x) = \sin x$.
Then,the derivative is $f'(x) = \cos x$.
Substituting these into the formula,we get:
$I = e^{x} \sin x + C$,where $C$ is an arbitrary constant.

Let $I = \int \frac{x e^{x}}{(1+x)^{2}} dx = \int e^{x} \left\{ \frac{x}{(1+x)^{2}} \right\} dx$
$= \int e^{x} \left\{ \frac{1+x-1}{(1+x)^{2}} \right\} dx$
$= \int e^{x} \left\{ \frac{1}{1+x} - \frac{1}{(1+x)^{2}} \right\} dx$
Let $f(x) = \frac{1}{1+x}$,then $f'(x) = -\frac{1}{(1+x)^{2}}$
$\Rightarrow \int \frac{x e^{x}}{(1+x)^{2}} dx = \int e^{x} \{f(x) + f'(x)\} dx$
It is known that $\int e^{x} \{f(x) + f'(x)\} dx = e^{x} f(x) + C$
$\therefore \int \frac{x e^{x}}{(1+x)^{2}} dx = \frac{e^{x}}{1+x} + C$
Where $C$ is an arbitrary constant.

We need to evaluate the integral $I = \int e^{x} \left( \frac{1+\sin x}{1+\cos x} \right) dx$.
Using trigonometric identities,$1+\sin x = \sin^{2} \frac{x}{2} + \cos^{2} \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2} = (\sin \frac{x}{2} + \cos \frac{x}{2})^{2}$ and $1+\cos x = 2 \cos^{2} \frac{x}{2}$.
Substituting these into the expression:
$I = \int e^{x} \left( \frac{(\sin \frac{x}{2} + \cos \frac{x}{2})^{2}}{2 \cos^{2} \frac{x}{2}} \right) dx$
$I = \int e^{x} \cdot \frac{1}{2} \left( \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} + 1 \right)^{2} dx$
$I = \frac{1}{2} \int e^{x} (\tan \frac{x}{2} + 1)^{2} dx$
$I = \frac{1}{2} \int e^{x} (\tan^{2} \frac{x}{2} + 1 + 2 \tan \frac{x}{2}) dx$
Since $\tan^{2} \frac{x}{2} + 1 = \sec^{2} \frac{x}{2}$,we have:
$I = \int e^{x} (\frac{1}{2} \sec^{2} \frac{x}{2} + \tan \frac{x}{2}) dx$
Let $f(x) = \tan \frac{x}{2}$. Then $f'(x) = \frac{1}{2} \sec^{2} \frac{x}{2}$.
Using the standard integral formula $\int e^{x} (f(x) + f'(x)) dx = e^{x} f(x) + C$,we get:
$I = e^{x} \tan \frac{x}{2} + C$.

(A) Let $I = \int e^{x} \left[ \frac{1}{x} - \frac{1}{x^{2}} \right] dx$.
We know the standard integral formula: $\int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + C$.
Here,let $f(x) = \frac{1}{x}$.
Then,the derivative is $f'(x) = -\frac{1}{x^{2}}$.
Substituting these into the formula,we get:
$I = e^{x} \left( \frac{1}{x} \right) + C = \frac{e^{x}}{x} + C$,where $C$ is an arbitrary constant.

We need to evaluate the integral $I = \int \frac{(x-3) e^{x}}{(x-1)^{3}} dx$.
Rewrite the numerator as $(x-1-2)$:
$I = \int e^{x} \left\{ \frac{x-1-2}{(x-1)^{3}} \right\} dx$
Split the fraction:
$I = \int e^{x} \left\{ \frac{x-1}{(x-1)^{3}} - \frac{2}{(x-1)^{3}} \right\} dx = \int e^{x} \left\{ \frac{1}{(x-1)^{2}} - \frac{2}{(x-1)^{3}} \right\} dx$
Let $f(x) = \frac{1}{(x-1)^{2}}$. Then,$f'(x) = -2(x-1)^{-3} = \frac{-2}{(x-1)^{3}}$.
Using the standard integral formula $\int e^{x} \{f(x) + f'(x)\} dx = e^{x} f(x) + C$:
$I = e^{x} \left( \frac{1}{(x-1)^{2}} \right) + C = \frac{e^{x}}{(x-1)^{2}} + C$,where $C$ is an arbitrary constant.

(D) We are given the integral $I = \int e^{x} \sec x(1+\tan x) d x$.
Distributing the term $\sec x$,we get:
$I = \int e^{x}(\sec x + \sec x \tan x) d x$.
Recall the standard integration formula: $\int e^{x} \{f(x) + f'(x)\} d x = e^{x} f(x) + C$.
Let $f(x) = \sec x$. Then,the derivative $f'(x) = \frac{d}{dx}(\sec x) = \sec x \tan x$.
Substituting these into the formula,we obtain:
$I = e^{x} \sec x + C$.
Thus,the correct option is $D$.

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 $1 + \cos 2x = 2 \cos^2 x$:
$I = \int \left( \frac{2 + 2 \sin x \cos x}{2 \cos^2 x} \right) e^x dx$
$I = \int \left( \frac{2(1 + \sin x \cos x)}{2 \cos^2 x} \right) e^x dx$
$I = \int \left( \frac{1}{\cos^2 x} + \frac{\sin x \cos x}{\cos^2 x} \right) e^x dx$
$I = \int (\sec^2 x + \tan x) e^x dx$
We know that $\int (f(x) + f'(x)) e^x dx = e^x f(x) + C$.
Here,let $f(x) = \tan x$,then $f'(x) = \sec^2 x$.
Therefore,$I = e^x \tan x + C$.

(D) Let $I = \int_{1}^{2} e^{x} x^{x} (2 + \log_{e} x) dx$.
We can rewrite the integrand as:
$I = \int_{1}^{2} e^{x} [x^{x} + x^{x}(1 + \log_{e} x)] dx$.
Let $f(x) = x^{x}$.
Then $f'(x) = \frac{d}{dx}(e^{x \log_{e} x}) = e^{x \log_{e} x} \cdot \frac{d}{dx}(x \log_{e} x) = x^{x} (1 \cdot \log_{e} x + x \cdot \frac{1}{x}) = x^{x} (1 + \log_{e} x)$.
Thus,the integral is of the form $\int_{1}^{2} e^{x} [f(x) + f'(x)] dx$.
The result of this integral is $[e^{x} f(x)]_{1}^{2}$.
Substituting $f(x) = x^{x}$:
$I = [e^{x} x^{x}]_{1}^{2} = (e^{2} \cdot 2^{2}) - (e^{1} \cdot 1^{1}) = 4e^{2} - e = e(4e - 1)$.

(A) We have $\int \frac{(x^{2}+1) e^{x}}{(x+1)^{2}} d x = \int \frac{(x^{2}-1+2) e^{x}}{(x+1)^{2}} d x$.
$= \int \left( \frac{(x-1)(x+1)}{(x+1)^{2}} + \frac{2}{(x+1)^{2}} \right) e^{x} d x = \int \left( \frac{x-1}{x+1} + \frac{2}{(x+1)^{2}} \right) e^{x} d x$.
This is in the form $\int (f(x) + f'(x)) e^{x} d x = f(x) e^{x} + C$,where $f(x) = \frac{x-1}{x+1}$.
We can write $f(x) = \frac{x+1-2}{x+1} = 1 - 2(x+1)^{-1}$.
Then $f'(x) = 2(x+1)^{-2}$.
$f''(x) = -4(x+1)^{-3}$.
$f'''(x) = 12(x+1)^{-4} = \frac{12}{(x+1)^{4}}$.
At $x = 1$,$f'''(1) = \frac{12}{(1+1)^{4}} = \frac{12}{16} = \frac{3}{4}$.

(C) We know that $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$.
Let $f(x) = \frac{x \sin^{-1} x}{\sqrt{1-x^2}}$.
Then $f'(x) = \frac{d}{dx} \left( \frac{x \sin^{-1} x}{\sqrt{1-x^2}} \right) = \frac{\sqrt{1-x^2} \cdot \frac{d}{dx}(x \sin^{-1} x) - x \sin^{-1} x \cdot \frac{d}{dx}(\sqrt{1-x^2})}{1-x^2}$
$= \frac{\sqrt{1-x^2} \left( \sin^{-1} x + \frac{x}{\sqrt{1-x^2}} \right) - x \sin^{-1} x \left( \frac{-x}{\sqrt{1-x^2}} \right)}{1-x^2}$
$= \frac{\sqrt{1-x^2} \sin^{-1} x + x + \frac{x^2 \sin^{-1} x}{\sqrt{1-x^2}}}{1-x^2} = \frac{(1-x^2) \sin^{-1} x + x \sqrt{1-x^2} + x^2 \sin^{-1} x}{(1-x^2)^{3/2}} = \frac{\sin^{-1} x}{(1-x^2)^{3/2}} + \frac{x}{1-x^2}$.
Thus,the integral is $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C = e^x \frac{x \sin^{-1} x}{\sqrt{1-x^2}} + C$.
So,$g(x) = \frac{x e^x \sin^{-1} x}{\sqrt{1-x^2}}$.
Evaluating at $x = \frac{1}{2}$: $g\left(\frac{1}{2}\right) = \frac{\frac{1}{2} e^{1/2} \sin^{-1}(1/2)}{\sqrt{1-(1/2)^2}} = \frac{\frac{1}{2} \sqrt{e} \cdot \frac{\pi}{6}}{\sqrt{3/4}} = \frac{\frac{\pi \sqrt{e}}{12}}{\frac{\sqrt{3}}{2}} = \frac{\pi \sqrt{e}}{6 \sqrt{3}} = \frac{\pi}{6} \sqrt{\frac{e}{3}}$.

(C) We use the standard result $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$.
Let $f(x) = \log \sin x$.
Then $f'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x$.
Thus,the integral becomes $\int_{\pi / 4}^{\pi / 2} e^x (f(x) + f'(x)) dx = [e^x f(x)]_{\pi / 4}^{\pi / 2}$.
Substituting the limits:
$= [e^x \log \sin x]_{\pi / 4}^{\pi / 2} = e^{\pi / 2} \log \sin(\pi / 2) - e^{\pi / 4} \log \sin(\pi / 4)$.
Since $\sin(\pi / 2) = 1$ and $\log(1) = 0$,the first term is $0$.
$= 0 - e^{\pi / 4} \log(1 / \sqrt{2}) = -e^{\pi / 4} \log(2^{-1/2})$.
$= -e^{\pi / 4} \cdot (-1/2) \log 2 = \frac{1}{2} e^{\pi / 4} \log 2$.

(A) Let $I = \int_1^{e} \frac{e^x}{x}(1+x \log x) d x$.
We can rewrite the integrand as:
$I = \int_1^{e} (\frac{e^x}{x} + e^x \log x) d x$.
Let $f(x) = e^x \log x$.
Then $f'(x) = e^x \log x + e^x \cdot \frac{1}{x} = e^x (\log x + \frac{1}{x})$.
This matches the integrand exactly.
Therefore,$\int (f(x) + f'(x)) d x = f(x) + C$.
So,$I = [e^x \log x]_1^e$.
Evaluating the limits:
$I = (e^e \log e) - (e^1 \log 1)$.
Since $\log e = 1$ and $\log 1 = 0$,
$I = (e^e \cdot 1) - (e \cdot 0) = e^e$.
The correct option is $A$.

(B) We are given the integral $I = \int_2^{e}\left[\frac{1}{\log x}-\frac{1}{(\log x)^2}\right] dx$.
Let $\log x = t$,then $x = e^t$ and $dx = e^t dt$.
When $x = 2$,$t = \log 2$. When $x = e$,$t = 1$.
Substituting these into the integral:
$I = \int_{\log 2}^{1} \left(\frac{1}{t} - \frac{1}{t^2}\right) e^t dt$.
We know that $\int e^t [f(t) + f'(t)] dt = e^t f(t) + C$.
Here,$f(t) = \frac{1}{t}$ and $f'(t) = -\frac{1}{t^2}$.
Therefore,$I = \left[ e^t \cdot \frac{1}{t} \right]_{\log 2}^{1}$.
$I = \left( e^1 \cdot \frac{1}{1} \right) - \left( e^{\log 2} \cdot \frac{1}{\log 2} \right)$.
$I = e - \frac{2}{\log 2}$.
Comparing this with $a + \frac{b}{\log 2}$,we get $a = e$ and $b = -2$.

(A) We need to evaluate the integral $I = \int e^x \left( \frac{1 - \sin x}{1 - \cos x} \right) dx$.
Using trigonometric identities,$1 - \sin x = 1 - 2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $1 - \cos x = 2 \sin^2 \frac{x}{2}$.
Substituting these into the integral:
$I = \int e^x \left( \frac{1 - 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^2 \frac{x}{2}} \right) dx$
$I = \int e^x \left( \frac{1}{2 \sin^2 \frac{x}{2}} - \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^2 \frac{x}{2}} \right) dx$
$I = \int e^x \left( \frac{1}{2} \operatorname{cosec}^2 \frac{x}{2} - \cot \frac{x}{2} \right) dx$.
Let $f(x) = -\cot \frac{x}{2}$. Then $f'(x) = -(-\operatorname{cosec}^2 \frac{x}{2}) \cdot \frac{1}{2} = \frac{1}{2} \operatorname{cosec}^2 \frac{x}{2}$.
Since the integral is of the form $\int e^x (f(x) + f'(x)) dx = e^x f(x) + c$,
$I = e^x (-\cot \frac{x}{2}) + c = -e^x \cot \frac{x}{2} + c$.

(C) Let $I = \int \frac{(\log x-1)^2}{\left\{1+(\log x)^2\right\}^2} dx$.
Substitute $\log x = t$,so $x = e^t$ and $dx = e^t dt$.
Then $I = \int \frac{(t-1)^2}{(1+t^2)^2} e^t dt = \int e^t \left( \frac{t^2-2t+1}{(1+t^2)^2} \right) dt$.
We can rewrite the integrand as $\int e^t \left( \frac{t^2+1-2t}{(1+t^2)^2} \right) dt = \int e^t \left( \frac{1}{1+t^2} - \frac{2t}{(1+t^2)^2} \right) dt$.
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,$I = e^t \left( \frac{1}{1+t^2} \right) + C$.
Substituting back $t = \log x$,we get $I = \frac{e^{\log x}}{1+(\log x)^2} + C = \frac{x}{1+(\log x)^2} + C$.

(D) We know that the integral of the form $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c$.
Here,let $f(x) = \frac{1}{x}$.
Then,$f'(x) = -\frac{1}{x^2}$.
Substituting these into the integral:
$I = \int e^x \left( \frac{1}{x} - \frac{1}{x^2} \right) dx = \int e^x \left( f(x) + f'(x) \right) dx$.
Using the formula,we get $I = e^x f(x) + c = \frac{e^x}{x} + c$.

(D) Let $t = \sin^{-1} x$,then $x = \sin t$ and $dx = \cos t \, dt$.
Substituting these into the integral:
$\int e^{\sin^{-1} x} \left( \frac{x}{\sqrt{1-x^2}} + 1 \right) dx = \int e^t \left( \frac{\sin t}{\cos t} + 1 \right) \cos t \, dt$
$= \int e^t (\sin t + \cos t) dt$.
Using the standard integral formula $\int e^t (f(t) + f'(t)) dt = e^t f(t) + c$,where $f(t) = \sin t$ and $f'(t) = \cos t$,we get:
$= e^t \sin t + c = e^{\sin^{-1} x} \cdot x + c$.

(B) Let $I = \int \frac{e^{\tan ^{-1} x}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] d x$.
Put $x = \tan t$,so $dx = \sec^2 t \, dt$.
Since $x > 0$,$t = \tan^{-1} x \in (0, \pi/2)$.
Then $\sec^{-1} \sqrt{1+x^2} = \sec^{-1}(\sec t) = t$ and $\cos^{-1}(\frac{1-x^2}{1+x^2}) = \cos^{-1}(\cos 2t) = 2t$.
Substituting these into the integral:
$I = \int \frac{e^t}{1+\tan^2 t} [t^2 + 2t] \sec^2 t \, dt = \int e^t (t^2 + 2t) \, dt$.
Using the formula $\int e^t (f(t) + f'(t)) \, dt = e^t f(t) + c$,where $f(t) = t^2$ and $f'(t) = 2t$:
$I = e^t \cdot t^2 + c = e^{\tan^{-1} x} (\tan^{-1} x)^2 + c$.

(A) Let $I = \int \frac{e^{\tan ^{-1} x}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] d x$.
Substitute $x = \tan t$,so $dx = \sec^2 t dt$.
Since $x > 0$,$t = \tan^{-1} x \in (0, \pi/2)$.
Note that $\sec^{-1} \sqrt{1+x^2} = \sec^{-1} \sqrt{1+\tan^2 t} = \sec^{-1} \sec t = t$.
Also,$\cos^{-1} \left(\frac{1-x^2}{1+x^2}\right) = \cos^{-1} \cos 2t = 2t$.
Substituting these into the integral:
$I = \int \frac{e^t}{1+\tan^2 t} [t^2 + 2t] \sec^2 t dt = \int e^t (t^2 + 2t) dt$.
Using the formula $\int e^t (f(t) + f'(t)) dt = e^t f(t) + c$,where $f(t) = t^2$ and $f'(t) = 2t$:
$I = e^t t^2 + c = e^{\tan^{-1} x} (\tan^{-1} x)^2 + c$.

(B) We are given the integral $I = \int \frac{x e^{2x}}{(1+2x)^2} dx$.
Let $2x = t$,then $2 dx = dt$,which implies $dx = \frac{dt}{2}$.
Substituting these into the integral:
$I = \int \frac{(t/2) e^t}{(1+t)^2} \cdot \frac{dt}{2} = \frac{1}{4} \int \frac{t e^t}{(1+t)^2} dt$.
We can rewrite the numerator $t$ as $(t+1-1)$:
$I = \frac{1}{4} \int e^t \left( \frac{t+1-1}{(1+t)^2} \right) dt = \frac{1}{4} \int e^t \left( \frac{1}{1+t} - \frac{1}{(1+t)^2} \right) dt$.
Using the standard integral formula $\int e^t (f(t) + f'(t)) dt = e^t f(t) + C$,where $f(t) = \frac{1}{1+t}$ and $f'(t) = -\frac{1}{(1+t)^2}$:
$I = \frac{1}{4} e^t \left( \frac{1}{1+t} \right) + C$.
Substituting $t = 2x$ back:
$I = \frac{e^{2x}}{4(1+2x)} + C$.

(A) We have the integral $I = \int e^{\tan x}(\sec^2 x + \sec^3 x \sin x) dx$.
Since $\sec^3 x \sin x = \sec^2 x \cdot \sec x \sin x = \sec^2 x \tan x$,we can rewrite the integral as:
$I = \int e^{\tan x}(\sec^2 x + \sec^2 x \tan x) dx = \int e^{\tan x} \sec^2 x (1 + \tan x) dx$.
Let $u = \tan x$,then $du = \sec^2 x dx$.
Substituting these into the integral,we get:
$I = \int e^u (1 + u) du = \int (e^u + u e^u) du$.
Using the standard integral form $\int e^u (f(u) + f'(u)) du = e^u f(u) + c$,where $f(u) = u$ and $f'(u) = 1$:
$I = e^u \cdot u + c = e^{\tan x} \tan x + c$.

(D) Let $I = \int [\sin(\log x) + \cos(\log x)] dx$.
Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$.
Substituting these into the integral,we get:
$I = \int (\sin t + \cos t) e^t dt$.
Using the standard integral formula $\int e^t [f(t) + f'(t)] dt = e^t f(t) + c$,where $f(t) = \sin t$ and $f'(t) = \cos t$.
Therefore,$I = e^t \sin t + c$.
Substituting back $t = \log x$,we get $I = x \sin(\log x) + c$.

(B) Let $\cos ^{-1} x = t$. Then $x = \cos t$ and $dx = -\sin t \ dt$. Also,$\frac{dx}{\sqrt{1-x^2}} = -dt$.
Substituting these into the integral:
$I = \int e^t \left[ \frac{\cos t - \sin t}{\sin t} \right] (-dt) = \int e^t \left[ \frac{\sin t - \cos t}{\sin t} \right] dt$
Wait,let us simplify the expression directly:
$I = \int e^{\cos ^{-1} x} \left( \frac{x}{\sqrt{1-x^2}} - 1 \right) dx$.
Let $t = \cos ^{-1} x$,so $x = \cos t$ and $dx = -\sin t \ dt$.
$I = \int e^t \left( \frac{\cos t}{\sin t} - 1 \right) (-\sin t \ dt) = \int e^t (\sin t - \cos t) dt$.
Using the formula $\int e^t (f(t) + f'(t)) dt = e^t f(t) + c$,where $f(t) = \sin t$ and $f'(t) = \cos t$,we get:
$I = e^t \sin t + c$.
Since $t = \cos ^{-1} x$,$\sin t = \sqrt{1 - \cos^2 t} = \sqrt{1 - x^2}$.
This does not match the options. Let us re-evaluate:
$I = \int e^t (\sin t - \cos t) dt = -\int e^t (\cos t - \sin t) dt = -e^t \cos t + c$.
Substituting back $t = \cos ^{-1} x$:
$I = -e^{\cos ^{-1} x} \cdot x + c = -x e^{\cos ^{-1} x} + c$.

(A) Let $I = \int \log x \cdot [\log (ex)]^{-2} dx$
Since $\log(ex) = \log e + \log x = 1 + \log x$,the integral becomes:
$I = \int \frac{\log x}{(1 + \log x)^{2}} dx$
Substitute $\log x = t$,which implies $x = e^{t}$ and $dx = e^{t} dt$.
$I = \int \frac{t}{(1 + t)^{2}} e^{t} dt$
We can rewrite the numerator as $(t + 1 - 1)$:
$I = \int e^{t} \left( \frac{t + 1 - 1}{(1 + t)^{2}} \right) dt$
$I = \int e^{t} \left( \frac{1}{1 + t} - \frac{1}{(1 + t)^{2}} \right) dt$
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 $f'(t) = -\frac{1}{(1 + t)^{2}}$:
$I = e^{t} \cdot \frac{1}{1 + t} + C$
Substituting back $t = \log x$ and $e^{t} = x$:
$I = \frac{x}{1 + \log x} + C$

(A) Given integral: $I = \int e^x \left[ \frac{2 + \sin 2x}{1 + \cos 2x} \right] dx$
Using trigonometric identities $\sin 2x = 2 \sin x \cos x$ and $1 + \cos 2x = 2 \cos^2 x$:
$I = \int e^x \left[ \frac{2 + 2 \sin x \cos x}{2 \cos^2 x} \right] dx$
$I = \int e^x \left[ \frac{2(1 + \sin x \cos x)}{2 \cos^2 x} \right] dx$
$I = \int e^x \left[ \frac{1}{\cos^2 x} + \frac{\sin x \cos x}{\cos^2 x} \right] dx$
$I = \int e^x (\sec^2 x + \tan x) dx$
Since $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$,where $f(x) = \tan x$ and $f'(x) = \sec^2 x$:
$I = e^x \tan x + C$

(B) Let $I = \int e^{\tan x}(\sec ^{2} x + \sec ^{3} x \sin x) d x$.
Since $\sec ^{3} x \sin x = \sec ^{2} x \cdot \sec x \sin x = \sec ^{2} x \tan x$,the integral becomes:
$I = \int e^{\tan x}(\sec ^{2} x + \sec ^{2} x \tan x) d x$.
$I = \int e^{\tan x}(1 + \tan x) \sec ^{2} x d x$.
Let $t = \tan x$,then $dt = \sec ^{2} x d x$.
Substituting these into the integral:
$I = \int e^{t}(1 + t) d t$.
$I = \int (e^{t} + t e^{t}) d t$.
Using the integration by parts formula $\int (f(t) + f'(t)) e^{t} d t = f(t) e^{t} + c$,where $f(t) = t$ and $f'(t) = 1$:
$I = t e^{t} + c$.
Substituting back $t = \tan x$:
$I = \tan x e^{\tan x} + c$.

(C) We want to evaluate the integral $I = \int e^x \frac{x-1}{(x+1)^3} \, dx$.
Rewrite the numerator as $(x+1) - 2$:
$I = \int e^x \frac{(x+1) - 2}{(x+1)^3} \, dx$
$I = \int e^x \left( \frac{x+1}{(x+1)^3} - \frac{2}{(x+1)^3} \right) \, dx$
$I = \int e^x \left( \frac{1}{(x+1)^2} - \frac{2}{(x+1)^3} \right) \, dx$
Recall the standard integral form $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + c$.
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,$I = \frac{e^x}{(x+1)^2} + c$.

(D) Let $I = \int e^{2x} \frac{\sin 2x \cos 2x - 1}{\sin^2 2x} \, dx$.
We can rewrite the integrand as:
$I = \int e^{2x} \left( \frac{\sin 2x \cos 2x}{\sin^2 2x} - \frac{1}{\sin^2 2x} \right) \, dx$
$I = \int e^{2x} (\cot 2x - \csc^2 2x) \, dx$.
Let $f(x) = \cot 2x$. Then $f'(x) = -\csc^2 2x \cdot 2 = -2 \csc^2 2x$.
This does not fit the form $\int e^{ax} (f(x) + \frac{f'(x)}{a}) \, dx$.
Let us rewrite the integral as $I = \int e^{2x} \cot 2x \, dx - \int e^{2x} \csc^2 2x \, dx$.
Using integration by parts on $\int e^{2x} \cot 2x \, dx$:
Let $u = \cot 2x$,$dv = e^{2x} \, dx$. Then $du = -2 \csc^2 2x \, dx$ and $v = \frac{1}{2} e^{2x}$.
$I = (\cot 2x \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} (-2 \csc^2 2x) \, dx) - \int e^{2x} \csc^2 2x \, dx$.
$I = \frac{1}{2} e^{2x} \cot 2x + \int e^{2x} \csc^2 2x \, dx - \int e^{2x} \csc^2 2x \, dx + c$.
$I = \frac{1}{2} e^{2x} \cot 2x + c$.

(D) We use the standard integral formula: $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c$.
First,rewrite the integrand: $\frac{x+5}{(x+6)^2} = \frac{(x+6) - 1}{(x+6)^2} = \frac{1}{x+6} - \frac{1}{(x+6)^2}$.
Let $f(x) = \frac{1}{x+6}$.
Then $f'(x) = -\frac{1}{(x+6)^2}$.
Substituting these into the integral: $\int e^x \left( \frac{1}{x+6} - \frac{1}{(x+6)^2} \right) dx = \int e^x [f(x) + f'(x)] dx$.
Applying the formula,we get $e^x f(x) + c = \frac{e^x}{x+6} + c$.

(A) We have $I = \int \left(\frac{x+2}{x+4}\right)^2 e^x \, dx$.
First,rewrite the integrand:
$I = \int \left(\frac{(x+4)-2}{x+4}\right)^2 e^x \, dx = \int \left(1 - \frac{2}{x+4}\right)^2 e^x \, dx$.
Expanding the square:
$I = \int \left(1 - \frac{4}{x+4} + \frac{4}{(x+4)^2}\right) e^x \, dx$.
Let $f(x) = 1 - \frac{4}{x+4} = \frac{x+4-4}{x+4} = \frac{x}{x+4}$.
Then $f'(x) = \frac{d}{dx} \left(1 - 4(x+4)^{-1}\right) = 0 - 4(-1)(x+4)^{-2} = \frac{4}{(x+4)^2}$.
Since the integral is of the form $\int e^x (f(x) + f'(x)) \, dx = e^x f(x) + c$,
$I = e^x \left(\frac{x}{x+4}\right) + c$.

(B) Let $I = \int \frac{(x-1) e^x}{(x+1)^3} \,d x$
We can rewrite the numerator as $(x+1-2)$:
$I = \int \frac{(x+1-2) e^x}{(x+1)^3} \,d x$
$I = \int e^x \left[ \frac{x+1}{(x+1)^3} - \frac{2}{(x+1)^3} \right] d x$
$I = \int e^x \left[ \frac{1}{(x+1)^2} - \frac{2}{(x+1)^3} \right] d x$
Let $f(x) = \frac{1}{(x+1)^2} = (x+1)^{-2}$.
Then $f'(x) = -2(x+1)^{-3} = -\frac{2}{(x+1)^3}$.
Using the standard integral formula $\int e^x [f(x) + f'(x)] \,d x = e^x f(x) + c$:
$I = e^x \left( \frac{1}{(x+1)^2} \right) + c = \frac{e^x}{(x+1)^2} + c$.

(C) Let $\sin \theta = t$. Then $\cos \theta \, d\theta = dt$.
Substituting these into the integral,we get:
$I = \int e^t (\log t + \operatorname{cosec}^2 \theta) \, dt$.
Note that $\operatorname{cosec}^2 \theta = \frac{1}{\sin^2 \theta} = \frac{1}{t^2}$.
So,$I = \int e^t (\log t + \frac{1}{t^2}) \, dt = \int e^t \log t \, dt + \int \frac{e^t}{t^2} \, dt$.
Using integration by parts for $\int e^t \log t \, dt$:
$\int e^t \log t \, dt = \log t \cdot e^t - \int \frac{e^t}{t} \, dt$.
For $\int \frac{e^t}{t^2} \, dt$,using integration by parts (taking $u = \frac{1}{t^2}$ and $dv = e^t \, dt$):
$\int \frac{e^t}{t^2} \, dt = \frac{e^t}{t^2} - \int e^t (-\frac{2}{t^3}) \, dt$ (This approach is complex).
Alternatively,observe the form $\int e^t (f(t) + f'(t)) \, dt = e^t f(t) + c$.
Let $f(t) = \log t - \frac{1}{t}$. Then $f'(t) = \frac{1}{t} + \frac{1}{t^2}$.
This does not match directly. Let's re-evaluate:
$I = \int e^t \log t \, dt + \int \frac{e^t}{t^2} \, dt = e^t \log t - \int \frac{e^t}{t} \, dt + \int \frac{e^t}{t^2} \, dt$.
Actually,$\int e^t (\log t - \frac{1}{t}) \, dt = e^t \log t - \int \frac{e^t}{t} \, dt + \int \frac{e^t}{t} \, dt = e^t \log t$.
Wait,the derivative of $e^t (\log t - \frac{1}{t})$ is $e^t (\log t - \frac{1}{t}) + e^t (\frac{1}{t} + \frac{1}{t^2}) = e^t (\log t + \frac{1}{t^2})$.
Thus,$I = e^t (\log t - \frac{1}{t}) + c = e^{\sin \theta} (\log \sin \theta - \operatorname{cosec} \theta) + c$.

(D) $I = \int \frac{(x-1) e^x}{(x+1)^3} \,dx$
$I = \int \left( \frac{x+1-2}{(x+1)^3} \right) e^x \,dx$
$I = \int \left[ \frac{x+1}{(x+1)^3} - \frac{2}{(x+1)^3} \right] e^x \,dx$
$I = \int \left[ \frac{1}{(x+1)^2} - \frac{2}{(x+1)^3} \right] e^x \,dx$
Let $f(x) = \frac{1}{(x+1)^2}$,then $f'(x) = -2(x+1)^{-3} = \frac{-2}{(x+1)^3}$.
Using the formula $\int e^x (f(x) + f'(x)) \,dx = e^x f(x) + C$,we get:
$I = e^x \left( \frac{1}{(x+1)^2} \right) + C = \frac{e^x}{(x+1)^2} + C$.

(D) Let $I = \int \frac{1+\sin (\log x)}{1+\cos (\log x)} d x$.
Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$.
Then,$I = \int \frac{1+\sin t}{1+\cos t} e^t dt$.
Using trigonometric identities $1+\cos t = 2\cos^2(t/2)$ and $\sin t = 2\sin(t/2)\cos(t/2)$,we get:
$I = \int \frac{\sin^2(t/2) + \cos^2(t/2) + 2\sin(t/2)\cos(t/2)}{2\cos^2(t/2)} e^t dt$.
$I = \frac{1}{2} \int (\tan^2(t/2) + 1 + 2\tan(t/2)) e^t dt$.
$I = \frac{1}{2} \int (\sec^2(t/2) + 2\tan(t/2)) e^t dt$.
Using the formula $\int e^t (f(t) + f'(t)) dt = e^t f(t) + c$,where $f(t) = \tan(t/2)$ and $f'(t) = \frac{1}{2}\sec^2(t/2)$:
$I = e^t \tan(t/2) + c$.
Substituting $t = \log x$ back,we get $I = x \tan \left(\frac{\log x}{2}\right) + c$.

(A) We need to evaluate the integral $I = \int \frac{x-3}{(x-1)^3} e^x \, dx$.
Rewrite the numerator as $(x-1) - 2$:
$I = \int \frac{(x-1) - 2}{(x-1)^3} e^x \, dx$
$I = \int \left[ \frac{x-1}{(x-1)^3} - \frac{2}{(x-1)^3} \right] e^x \, dx$
$I = \int e^x \left[ \frac{1}{(x-1)^2} - \frac{2}{(x-1)^3} \right] \, dx$
Let $f(x) = \frac{1}{(x-1)^2} = (x-1)^{-2}$.
Then,$f'(x) = -2(x-1)^{-3} = -\frac{2}{(x-1)^3}$.
Using the standard integral formula $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + c$,we get:
$I = e^x \left( \frac{1}{(x-1)^2} \right) + c$.

(A) We need to evaluate the integral $I = \int e^x \left( \frac{x^2+4x+4}{(x+4)^2} \right) dx$.
First,rewrite the numerator: $x^2+4x+4 = x(x+4) + 4$.
So,the integral becomes:
$I = \int e^x \left( \frac{x(x+4) + 4}{(x+4)^2} \right) dx$
$I = \int e^x \left( \frac{x}{x+4} + \frac{4}{(x+4)^2} \right) dx$.
Let $f(x) = \frac{x}{x+4}$.
Then $f'(x) = \frac{(x+4)(1) - x(1)}{(x+4)^2} = \frac{x+4-x}{(x+4)^2} = \frac{4}{(x+4)^2}$.
Since the integral is of the form $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c$,
we have $I = e^x \left( \frac{x}{x+4} \right) + c$.

(B) Let $I = \int \frac{e^{x}}{\sqrt{x}}(1+2x) dx$.
We can rewrite the integrand as:
$I = \int e^{x} \left( \frac{1}{\sqrt{x}} + \frac{2x}{\sqrt{x}} \right) dx = \int e^{x} \left( \frac{1}{\sqrt{x}} + 2\sqrt{x} \right) dx$.
Rearranging the terms:
$I = \int e^{x} \left( 2\sqrt{x} + \frac{1}{\sqrt{x}} \right) dx$.
Let $f(x) = 2\sqrt{x}$. Then $f'(x) = 2 \cdot \frac{1}{2\sqrt{x}} = \frac{1}{\sqrt{x}}$.
Using the standard integral formula $\int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + c$,we get:
$I = e^{x} (2\sqrt{x}) + c = 2\sqrt{x} e^{x} + c$.

(B) Let $I = \int e^{\tan ^{-1} x} \left(1 + \frac{x}{1 + x^2} \right) dx$.
Substitute $t = \tan ^{-1} x$,which implies $x = \tan t$ and $dx = (1 + x^2) dt = \sec^2 t dt$.
Then,$I = \int e^t \left(1 + \frac{\tan t}{1 + \tan^2 t} \right) \sec^2 t dt$.
Since $1 + \tan^2 t = \sec^2 t$,the expression becomes $I = \int e^t \left(1 + \frac{\tan t}{\sec^2 t} \right) \sec^2 t dt$.
$I = \int e^t (\sec^2 t + \tan t) dt$.
Using the standard integral formula $\int e^t [f(t) + f'(t)] dt = e^t f(t) + c$,where $f(t) = \tan t$ and $f'(t) = \sec^2 t$.
Thus,$I = e^t \tan t + c$.
Substituting back $t = \tan ^{-1} x$,we get $I = x e^{\tan ^{-1} x} + c$.

(B) Let $I = \int \left[ \frac{\log x - 1}{1 + (\log x)^2} \right]^2 dx$.
Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$.
The integral becomes $I = \int \left[ \frac{t - 1}{1 + t^2} \right]^2 e^t dt = \int e^t \left[ \frac{t^2 - 2t + 1}{(1 + t^2)^2} \right] dt$.
We can rewrite the numerator as $(t^2 + 1) - 2t$.
So,$I = \int e^t \left[ \frac{t^2 + 1}{(1 + t^2)^2} - \frac{2t}{(1 + t^2)^2} \right] dt = \int e^t \left[ \frac{1}{1 + t^2} - \frac{2t}{(1 + t^2)^2} \right] dt$.
This is in the form $\int e^t [f(t) + f'(t)] dt = e^t f(t) + c$,where $f(t) = \frac{1}{1 + t^2}$.
Thus,$I = e^t \left( \frac{1}{1 + t^2} \right) + c$.
Substituting back $t = \log x$,we get $I = \frac{x}{1 + (\log x)^2} + c$.

(A) Let $I = \int \log x \cdot (\log x + 2) dx$.
Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$.
Then,$I = \int t(t + 2) e^t dt = \int (t^2 + 2t) e^t dt$.
Recall the standard integral formula $\int e^t [f(t) + f'(t)] dt = e^t f(t) + c$.
Here,let $f(t) = t^2$,then $f'(t) = 2t$.
Thus,$I = e^t (t^2) + c$.
Substituting back $t = \log x$,we get $I = x(\log x)^2 + c$.

(B) We know the standard integral form $\int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + c$.
Given the integral $\int e^{x} \sec x(1+\tan x) dx$.
Distributing $\sec x$,we get $\int e^{x} (\sec x + \sec x \tan x) dx$.
Let $f(x) = \sec x$. Then $f'(x) = \sec x \tan x$.
Substituting this into the standard form,we get $e^{x} \sec x + c$.

(D) Let $I = \int e^{x} \left[ \frac{1+\sin x}{1+\cos x} \right] dx$.
Using the identities $1+\cos x = 2\cos^2 \frac{x}{2}$ and $\sin x = 2\sin \frac{x}{2} \cos \frac{x}{2}$,we get:
$I = \int e^{x} \left[ \frac{1 + 2\sin \frac{x}{2} \cos \frac{x}{2}}{2\cos^2 \frac{x}{2}} \right] dx$
$I = \int e^{x} \left[ \frac{1}{2\cos^2 \frac{x}{2}} + \frac{2\sin \frac{x}{2} \cos \frac{x}{2}}{2\cos^2 \frac{x}{2}} \right] dx$
$I = \int e^{x} \left[ \frac{1}{2} \sec^2 \frac{x}{2} + \tan \frac{x}{2} \right] dx$.
We know that $\int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + C$.
Here,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 $e^{x} \tan \frac{x}{2} + C$.

(C) We know that $\int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + c$.
The given integral is $\int e^{x} \left(\frac{x-1}{x^{2}}\right) dx$.
This can be rewritten as $\int e^{x} \left(\frac{1}{x} - \frac{1}{x^{2}}\right) dx$.
Here,let $f(x) = \frac{1}{x}$,then $f'(x) = -\frac{1}{x^{2}}$.
Applying the formula,we get $\int e^{x} \left(\frac{1}{x} - \frac{1}{x^{2}}\right) dx = e^{x} \left(\frac{1}{x}\right) + c = \frac{e^{x}}{x} + c$.

(D) We know that $\int [f(x) + x f'(x)] dx = x f(x) + c$.
Let $I = \int \left(1+x-\frac{1}{x}\right) e^{x+\frac{1}{x}} dx$.
We can rewrite the integrand as:
$I = \int \left( x \left(1 - \frac{1}{x^2}\right) + 1 \right) e^{x+\frac{1}{x}} dx$.
Let $f(x) = e^{x+\frac{1}{x}}$. Then $f'(x) = e^{x+\frac{1}{x}} \cdot \frac{d}{dx} \left(x + \frac{1}{x}\right) = e^{x+\frac{1}{x}} \left(1 - \frac{1}{x^2}\right)$.
Substituting this into the integral,we get:
$I = \int [x f'(x) + f(x)] dx$.
Using the formula $\int [x f'(x) + f(x)] dx = x f(x) + c$,we obtain:
$I = x e^{x+\frac{1}{x}} + c$.

(C) We know that $1 + \cot^2 x = \operatorname{cosec}^2 x$.
Substituting this into the integral,we get:
$\int e^x(1 - \cot x + \cot^2 x) dx = \int e^x(\operatorname{cosec}^2 x - \cot x) dx$.
Let $f(x) = -\cot x$.
Then $f'(x) = -(-\operatorname{cosec}^2 x) = \operatorname{cosec}^2 x$.
Using the standard formula $\int e^x[f(x) + f'(x)] dx = e^x f(x) + c$,we have:
$\int e^x(-\cot x + \operatorname{cosec}^2 x) dx = e^x(-\cot x) + c = -e^x \cdot \cot x + c$.

(A) 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 $1+\cos x = 2 \cos^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$:
$I = \int e^x \left( \frac{1 + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \right) dx$
$I = \int e^x \left( \frac{1}{2 \cos^2 \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \right) dx$
$I = \int e^x \left( \frac{1}{2} \sec^2 \frac{x}{2} + \tan \frac{x}{2} \right) dx$
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 $\int e^x [f(x) + f'(x)] dx$,the result is $e^x f(x) + c$.
Therefore,$I = e^x \tan \frac{x}{2} + c$.

(B) Let $I = \int \left[\frac{1-\log x}{1+(\log x)^{2}}\right]^{2} dx$.
Put $\log x = t$,then $x = e^{t}$,so $dx = e^{t} dt$.
Substituting these into the integral,we get $I = \int e^{t} \left[\frac{1-t}{1+t^{2}}\right]^{2} dt$.
Expanding the square,$I = \int e^{t} \frac{1+t^{2}-2t}{(1+t^{2})^{2}} dt = \int e^{t} \left[\frac{1}{1+t^{2}} - \frac{2t}{(1+t^{2})^{2}}\right] dt$.
Using the standard integral 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,$I = e^{t} \left(\frac{1}{1+t^{2}}\right) + c$.
Substituting back $t = \log x$,we get $I = \frac{x}{1+(\log x)^{2}} + c$.

(A) We know that $\int e^{x}[f(x) + f'(x)] dx = e^{x}f(x) + C$.
Given integral is $I = \int e^{x} \left( \frac{1-x}{1+x^{2}} \right)^{2} dx$.
Expanding the numerator: $I = \int e^{x} \frac{1 + x^{2} - 2x}{(1+x^{2})^{2}} dx$.
Splitting the fraction: $I = \int e^{x} \left[ \frac{1+x^{2}}{(1+x^{2})^{2}} - \frac{2x}{(1+x^{2})^{2}} \right] dx$.
$I = \int e^{x} \left[ \frac{1}{1+x^{2}} - \frac{2x}{(1+x^{2})^{2}} \right] dx$.
Let $f(x) = \frac{1}{1+x^{2}}$. Then $f'(x) = -\frac{1}{(1+x^{2})^{2}} \cdot (2x) = -\frac{2x}{(1+x^{2})^{2}}$.
Thus,$I = \int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + C$.
Therefore,$I = \frac{e^{x}}{1+x^{2}} + C$.

(C) Let $I = \int [\sin (\log x) + \cos (\log x)] \, dx$.
We know that $\frac{d}{dx} [x \sin (\log x)] = \sin (\log x) \cdot \frac{d}{dx}(x) + x \cdot \cos (\log x) \cdot \frac{d}{dx}(\log x)$.
$= \sin (\log x) \cdot 1 + x \cdot \cos (\log x) \cdot \frac{1}{x}$.
$= \sin (\log x) + \cos (\log x)$.
Therefore,$\int [\sin (\log x) + \cos (\log x)] \, dx = \int \frac{d}{dx} [x \sin (\log x)] \, dx$.
$= x \sin (\log x) + c$.

(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 $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c = e^x \tan \frac{x}{2} + c$.