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

(A) Let $I = \int \frac{3^x(x \log 3 - 1)}{x^2} dx$.
We can rewrite the integrand as:
$I = \int \left( \frac{x \cdot 3^x \log 3}{x^2} - \frac{3^x}{x^2} \right) dx$
$I = \int \left( \frac{3^x \log 3}{x} - \frac{3^x}{x^2} \right) dx$.
Recall the derivative rule for a quotient or product. Let $f(x) = \frac{3^x}{x}$.
Then $f'(x) = \frac{x \cdot \frac{d}{dx}(3^x) - 3^x \cdot \frac{d}{dx}(x)}{x^2}$.
Since $\frac{d}{dx}(3^x) = 3^x \log 3$,we have:
$f'(x) = \frac{x \cdot 3^x \log 3 - 3^x \cdot 1}{x^2} = \frac{3^x(x \log 3 - 1)}{x^2}$.
Thus,$\int f'(x) dx = f(x) + c$.
Therefore,$I = \frac{3^x}{x} + c$.

(C) Let $I = \int e^{-2 x}(\tan 2 x - 2 \sec^2 2 x \tan 2 x) dx$.
Consider the function $f(x) = \tan 2 x$.
Then $f'(x) = 2 \sec^2 2 x$.
The integral is of the form $\int e^{ax} (f(x) + \frac{1}{a} f'(x)) dx$ is not directly applicable here due to the term $\tan 2 x$.
Let $2x = t$,then $2 dx = dt$,so $dx = \frac{1}{2} dt$.
$I = \frac{1}{2} \int e^{-t} (\tan t - 2 \sec^2 t \tan t) dt$.
Let $f(t) = \tan t - \sec^2 t$.
Then $f'(t) = \sec^2 t - 2 \sec^2 t \tan t$.
Thus,$I = \frac{1}{2} \int e^{-t} (f'(t) - f(t)) dt$.
Using the formula $\int e^{kt} (f'(t) + k f(t)) dt = e^{kt} f(t) + C$,with $k = -1$:
$I = \frac{1}{2} (-e^{-t} f(t)) + C = -\frac{1}{2} e^{-t} (\tan t - \sec^2 t) + C$.
Substituting $t = 2x$:
$I = -\frac{1}{2} e^{-2 x} (\tan 2 x - \sec^2 2 x) + C = \frac{1}{2} e^{-2 x} (\sec^2 2 x - \tan 2 x) + C$.

(B) We are given the integral $I = \int e^x(\sin^2 2x - 8 \cos 4x) dx$.
Using the identity $\sin^2 2x = \frac{1 - \cos 4x}{2}$,we substitute this into the integral:
$I = \int e^x(\frac{1 - \cos 4x}{2} - 8 \cos 4x) dx = \int e^x(\frac{1}{2} - \frac{1}{2} \cos 4x - 8 \cos 4x) dx = \int e^x(\frac{1}{2} - \frac{17}{2} \cos 4x) dx$.
This simplifies to $I = \frac{1}{2} \int e^x dx - \frac{17}{2} \int e^x \cos 4x dx$.
Using the standard formula $\int e^{ax} \cos(bx) dx = \frac{e^{ax}}{a^2 + b^2} (a \cos bx + b \sin bx)$,we have $\int e^x \cos 4x dx = \frac{e^x}{1^2 + 4^2} (\cos 4x + 4 \sin 4x) = \frac{e^x}{17} (\cos 4x + 4 \sin 4x)$.
Substituting this back:
$I = \frac{1}{2} e^x - \frac{17}{2} \cdot \frac{e^x}{17} (\cos 4x + 4 \sin 4x) + c = e^x (\frac{1}{2} - \frac{1}{2} \cos 4x - 2 \sin 4x) + c$.
Thus,$f(x) = \frac{1}{2} - \frac{1}{2} \cos 4x - 2 \sin 4x$.
Evaluating at $x = \frac{\pi}{4}$:
$f(\frac{\pi}{4}) = \frac{1}{2} - \frac{1}{2} \cos(\pi) - 2 \sin(\pi) = \frac{1}{2} - \frac{1}{2}(-1) - 2(0) = \frac{1}{2} + \frac{1}{2} = 1$.

(D) We know that $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$.
Let $f(x) = \frac{x-4}{(x-1)^4}$.
Then $f'(x) = \frac{(x-1)^4(1) - (x-4) \cdot 4(x-1)^3}{(x-1)^8} = \frac{(x-1) - 4(x-4)}{(x-1)^5} = \frac{x-1-4x+16}{(x-1)^5} = \frac{-3x+15}{(x-1)^5} = \frac{-3(x-5)}{(x-1)^5}$.
Now,consider the integral $I = \int e^x \left( \frac{x^2-8x+19}{(x-1)^5} \right) dx$.
We can rewrite the numerator as $x^2-8x+19 = (x-4)(x-1) - 3(x-5)$.
So,$I = \int e^x \left( \frac{x-4}{(x-1)^4} - \frac{3(x-5)}{(x-1)^5} \right) dx = \int e^x [f(x) + f'(x)] dx = e^x f(x) + C = \frac{e^x(x-4)}{(x-1)^4} + C$.
Comparing this with $\frac{e^x(lx+m)}{(x-1)^4} + C$,we get $l=1$ and $m=-4$.
Therefore,$4l+m = 4(1) + (-4) = 4-4 = 0$.

(B) We are given that $\int e^{2x} f^{\prime}(x) dx = g(x)$.
Let $I = \int (e^{2x} f(x) + e^{2x} f^{\prime}(x)) dx$.
This can be split into two integrals: $I = \int e^{2x} f(x) dx + \int e^{2x} f^{\prime}(x) dx$.
Using integration by parts for the first integral $\int e^{2x} f(x) dx$ (taking $f(x)$ as the first function and $e^{2x}$ as the second function):
$\int e^{2x} f(x) dx = f(x) \int e^{2x} dx - \int (f^{\prime}(x) \int e^{2x} dx) dx = \frac{1}{2} f(x) e^{2x} - \frac{1}{2} \int e^{2x} f^{\prime}(x) dx$.
Substituting this back into the expression for $I$:
$I = [\frac{1}{2} f(x) e^{2x} - \frac{1}{2} \int e^{2x} f^{\prime}(x) dx] + \int e^{2x} f^{\prime}(x) dx$.
$I = \frac{1}{2} e^{2x} f(x) + \frac{1}{2} \int e^{2x} f^{\prime}(x) dx$.
Since $\int e^{2x} f^{\prime}(x) dx = g(x)$,we have:
$I = \frac{1}{2} e^{2x} f(x) + \frac{1}{2} g(x) + C = \frac{1}{2} [e^{2x} f(x) + g(x)] + C$.

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

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

(D) We know that $\int e^x(g(x) + g'(x)) dx = e^x g(x) + c$.
Given $\int e^x(x^3+x^2-x+4) dx = e^x f(x) + c$,we have $f(x) + f'(x) = x^3+x^2-x+4$.
Let $f(x) = ax^3 + bx^2 + cx + d$. Then $f'(x) = 3ax^2 + 2bx + c$.
Substituting these into the equation: $ax^3 + (a+b)x^2 + (b+c)x + (c+d) = x^3 + x^2 - x + 4$.
Comparing coefficients:
$a = 1$.
$a + b = 1 \Rightarrow 1 + b = 1 \Rightarrow b = 0$.
$b + c = -1 \Rightarrow 0 + c = -1 \Rightarrow c = -1$.
$c + d = 4 \Rightarrow -1 + d = 4 \Rightarrow d = 5$.
Thus,$f(x) = x^3 - x + 5$.
Calculating $f(1) = 1^3 - 1 + 5 = 5$.
Wait,let us re-evaluate the derivative approach:
If $f(x) = x^3 - 2x^2 + 3x + 1$,then $f'(x) = 3x^2 - 4x + 3$.
$f(x) + f'(x) = x^3 - 2x^2 + 3x + 1 + 3x^2 - 4x + 3 = x^3 + x^2 - x + 4$.
This matches the integrand.
Therefore,$f(1) = 1^3 - 2(1)^2 + 3(1) + 1 = 1 - 2 + 3 + 1 = 3$.

(C) We are given the integral $I = \int e^x \left( \frac{1 - \sin x}{1 - \cos x} \right) dx$.
Using trigonometric identities $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $1 - \cos x = 2 \sin^2 \frac{x}{2}$,we get:
$I = \int e^x \left( \frac{1 - 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^2 \frac{x}{2}} \right) dx$
$I = \int e^x \left( \frac{1}{2 \sin^2 \frac{x}{2}} - \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^2 \frac{x}{2}} \right) dx$
$I = \int e^x \left( \frac{1}{2} \operatorname{cosec}^2 \frac{x}{2} - \cot \frac{x}{2} \right) dx$
Let $g(x) = -\cot \frac{x}{2}$. Then $g'(x) = -(-\operatorname{cosec}^2 \frac{x}{2} \cdot \frac{1}{2}) = \frac{1}{2} \operatorname{cosec}^2 \frac{x}{2}$.
Since the integral is of the form $\int e^x (g(x) + g'(x)) dx = e^x g(x) + c$,we have:
$I = e^x \left( -\cot \frac{x}{2} \right) + c = -e^x \cot \frac{x}{2} + c$.
Thus,$f(x) = -e^x \cot \left( \frac{x}{2} \right)$.

(C) Let $I = \int 2^{x} [f^{\prime}(x) + f(x) \log 2] \, dx$.
We know that the derivative of the product of two functions is given by the product rule: $\frac{d}{dx} [u(x)v(x)] = u(x)v^{\prime}(x) + v(x)u^{\prime}(x)$.
Consider the function $g(x) = 2^{x} f(x)$.
Differentiating $g(x)$ with respect to $x$ using the product rule:
$g^{\prime}(x) = \frac{d}{dx}(2^{x}) \cdot f(x) + 2^{x} \cdot \frac{d}{dx}(f(x))$
$g^{\prime}(x) = 2^{x} \log 2 \cdot f(x) + 2^{x} f^{\prime}(x)$
$g^{\prime}(x) = 2^{x} [f^{\prime}(x) + f(x) \log 2]$.
Since the integrand is exactly the derivative of $g(x)$,we have:
$I = \int g^{\prime}(x) \, dx = g(x) + C = 2^{x} f(x) + C$.

(A) Given the integral: $\int e^{\sin x} \left( \frac{x \cos^3 x - \sin x}{\cos^2 x} \right) dx = e^{\sin x} f(x) + c$.
Simplify the integrand: $\frac{x \cos^3 x - \sin x}{\cos^2 x} = x \cos x - \frac{\sin x}{\cos^2 x} = x \cos x - \sec x \tan x$.
So the integral becomes: $\int e^{\sin x} (x \cos x - \sec x \tan x) dx$.
We can rewrite this as: $\int [e^{\sin x} \cos x \cdot x - e^{\sin x} \sec x \tan x] dx$.
Notice that $\frac{d}{dx} [e^{\sin x} (x - \sec x)] = e^{\sin x} \cos x (x - \sec x) + e^{\sin x} (1 - \sec x \tan x) = e^{\sin x} (x \cos x - \sec x \cos x + 1 - \sec x \tan x) = e^{\sin x} (x \cos x - 1 + 1 - \sec x \tan x) = e^{\sin x} (x \cos x - \sec x \tan x)$.
Thus,$\int e^{\sin x} (x \cos x - \sec x \tan x) dx = e^{\sin x} (x - \sec x) + c$.
Comparing this with $e^{\sin x} f(x) + c$,we get $f(x) = x - \sec x$.

(B) Let $I = \int 2^x (f^{\prime}(x) + f(x) \log 2) \, dx$.
We can rewrite the integrand as:
$I = \int (2^x f^{\prime}(x) + 2^x \log 2 \cdot f(x)) \, dx$.
Recall the product rule for differentiation: $\frac{d}{dx} [u(x)v(x)] = u(x)v^{\prime}(x) + v(x)u^{\prime}(x)$.
Let $u(x) = 2^x$ and $v(x) = f(x)$.
Then $u^{\prime}(x) = \frac{d}{dx} (2^x) = 2^x \log 2$.
Thus,the integrand is of the form $u(x)v^{\prime}(x) + v(x)u^{\prime}(x)$,which is $\frac{d}{dx} [2^x f(x)]$.
Therefore,$I = \int \frac{d}{dx} [2^x f(x)] \, dx = 2^x f(x) + C$.

(C) We know the standard integral formula: $\int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + C$.
Given the integral: $I = \int e^{x} \left(\frac{2}{x} - \frac{2}{x^2}\right) dx$.
Let $f(x) = \frac{2}{x}$.
Then,$f'(x) = \frac{d}{dx} (2x^{-1}) = -2x^{-2} = -\frac{2}{x^2}$.
Substituting these into the formula: $I = \int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + C$.
Therefore,$I = e^{x} \left(\frac{2}{x}\right) + C = \frac{2 e^{x}}{x} + C$.

(C) We know that $\int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + C$.
Let $f(x) = \log_{e} x$. Then $f'(x) = \frac{1}{x}$.
The integral can be rewritten as:
$I = \int_{1}^{2} e^{x} \left( \log_{e} x + \frac{1}{x} + 1 \right) dx$
$I = \int_{1}^{2} e^{x} \left( (\log_{e} x + \frac{1}{x}) + 1 \right) dx$
This does not directly fit the form. Let us expand:
$I = \int_{1}^{2} e^{x} \log_{e} x dx + \int_{1}^{2} e^{x} dx + \int_{1}^{2} \frac{e^{x}}{x} dx$
Using integration by parts for $\int e^{x} \log_{e} x dx$:
Let $u = \log_{e} x$,$dv = e^{x} dx$. Then $du = \frac{1}{x} dx$,$v = e^{x}$.
$\int e^{x} \log_{e} x dx = e^{x} \log_{e} x - \int \frac{e^{x}}{x} dx$.
Substituting this back into $I$:
$I = [e^{x} \log_{e} x - \int \frac{e^{x}}{x} dx] + [e^{x}]_{1}^{2} + \int_{1}^{2} \frac{e^{x}}{x} dx$
$I = [e^{x} \log_{e} x]_{1}^{2} + [e^{x}]_{1}^{2}$
$I = (e^{2} \log_{e} 2 - e^{1} \log_{e} 1) + (e^{2} - e^{1})$
Since $\log_{e} 1 = 0$:
$I = e^{2} \log_{e} 2 + e^{2} - e$
$I = e^{2}(1 + \log_{e} 2) - e$.

(A) Let $\ln t = x$,then $t = e^x$ and $dt = e^x dx$. Substituting these into the integral:
$f(t) = \int \frac{1 - \sin x}{1 - \cos x} e^x dx = \int \frac{1 - 2\sin(x/2)\cos(x/2)}{2\sin^2(x/2)} e^x dx$
$f(t) = \int \left( \frac{1}{2}\csc^2(x/2) - \cot(x/2) \right) e^x dx$
Using the identity $\int e^x (g(x) + g'(x)) dx = e^x g(x) + C$,where $g(x) = -\cot(x/2)$ and $g'(x) = \frac{1}{2}\csc^2(x/2)$:
$f(t) = -e^x \cot(x/2) + C = -t \cot(\frac{\ln t}{2}) + C$
Given $f(e^{\pi/2}) = -e^{\pi/2} \cot(\pi/4) + C = -e^{\pi/2} + C = -e^{\pi/2}$,so $C = 0$.
Thus,$f(t) = -t \cot(\frac{\ln t}{2})$.
$f(e^{\pi/4}) = -e^{\pi/4} \cot(\pi/8) = -e^{\pi/4} (\sqrt{2} + 1)$.
Comparing with $f(e^{\pi/4}) = \alpha e^{\pi/4}$,we get $\alpha = -(1 + \sqrt{2}) = -1 - \sqrt{2}$.

(A) We have $f(x) = \int e^x \left( \frac{2-x^2}{\sqrt{1+x}(1-x)^{3/2}} \right) dx$.
Rewrite the numerator as $(1-x^2) + 1$:
$f(x) = \int e^x \left( \frac{1-x^2}{\sqrt{1+x}(1-x)^{3/2}} + \frac{1}{\sqrt{1+x}(1-x)^{3/2}} \right) dx$.
Simplify the first term: $\frac{(1-x)(1+x)}{\sqrt{1+x}(1-x)^{3/2}} = \frac{\sqrt{1+x}}{\sqrt{1-x}} = \sqrt{\frac{1+x}{1-x}}$.
Let $g(x) = \sqrt{\frac{1+x}{1-x}}$. Then $g'(x) = \frac{1}{2} \left( \frac{1+x}{1-x} \right)^{-1/2} \cdot \frac{(1-x)(1) - (1+x)(-1)}{(1-x)^2} = \frac{1}{2} \sqrt{\frac{1-x}{1+x}} \cdot \frac{2}{(1-x)^2} = \frac{1}{\sqrt{1+x}(1-x)^{3/2}}$.
Thus,the integral is of the form $\int e^x (g(x) + g'(x)) dx = e^x g(x) + C$.
So,$f(x) = e^x \sqrt{\frac{1+x}{1-x}} + C$.
Given $f(0) = 0$,we have $0 = e^0 \sqrt{\frac{1}{1}} + C \implies 0 = 1 + C \implies C = -1$.
Therefore,$f(x) = e^x \sqrt{\frac{1+x}{1-x}} - 1$.
For $x = \frac{1}{2}$,$f(\frac{1}{2}) = e^{1/2} \sqrt{\frac{1+1/2}{1-1/2}} - 1 = \sqrt{e} \sqrt{\frac{3/2}{1/2}} - 1 = \sqrt{e} \sqrt{3} - 1 = \sqrt{3e} - 1$.

(C) Let $u = \tan^{-1} x$. Then $x = \tan u$ and $dx = \sec^2 u \, du = (1+x^2) \, du$.
Substituting these into the integral:
$I = \int e^u \left( \frac{1+x+x^2}{1+x^2} \right) (1+x^2) \, du = \int e^u (1+x+x^2) \, du$.
Since $x = \tan u$,we have $1+x^2 = \sec^2 u$.
So,$I = \int e^u (1 + \tan u + \tan^2 u) \, du = \int e^u (\sec^2 u + \tan u) \, du$.
We know that $\frac{d}{du}(\tan u) = \sec^2 u$.
Using the standard integral formula $\int e^u (f(u) + f'(u)) \, du = e^u f(u) + C$,where $f(u) = \tan u$ and $f'(u) = \sec^2 u$,we get:
$I = e^u \tan u + C = e^{\tan^{-1} x} \cdot x + C$.

(B) The given integral is $I = \int e^x \left( \frac{1-x}{1+x^2} \right)^2 dx$.
Expanding the numerator,we get $I = \int e^x \frac{1-2x+x^2}{(1+x^2)^2} dx$.
This can be rewritten as $I = \int e^x \left[ \frac{1+x^2}{(1+x^2)^2} - \frac{2x}{(1+x^2)^2} \right] dx$.
$I = \int e^x \left[ \frac{1}{1+x^2} - \frac{2x}{(1+x^2)^2} \right] dx$.
We know the standard integral formula $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$.
Here,let $f(x) = \frac{1}{1+x^2}$.
Then,$f'(x) = \frac{d}{dx} (1+x^2)^{-1} = -1(1+x^2)^{-2} \cdot (2x) = -\frac{2x}{(1+x^2)^2}$.
Since the integrand is in the form $e^x [f(x) + f'(x)]$,the solution is $e^x f(x) + C = \frac{e^x}{1+x^2} + C$.