Integration of rational function by using partial fractions, Questions in English

(C) Step $1$: Factorize the denominator: $x^2 - 3x + 2 = (x - 2)(x - 1)$.
Step $2$: Use partial fractions: $\frac{3x + 7}{(x - 2)(x - 1)} = \frac{A}{x - 2} + \frac{B}{x - 1}$.
Step $3$: Solve for $A$ and $B$: $3x + 7 = A(x - 1) + B(x - 2)$.
For $x = 2$: $3(2) + 7 = A(2 - 1) \implies A = 13$.
For $x = 1$: $3(1) + 7 = B(1 - 2) \implies 10 = -B \implies B = -10$.
Step $4$: Integrate: $\int (\frac{13}{x - 2} - \frac{10}{x - 1}) dx = 13 \log |x - 2| - 10 \log |x - 1| + c$.
Step $5$: Compare with $m \log |x - 2| + n \log |x - 1| + c$, we get $m = 13$ and $n = -10$.
Step $6$: Calculate $m + n = 13 + (-10) = 3$.

(B) We have $I = \int \frac{\sin x}{\sin 4x} dx = \int \frac{\sin x}{2 \sin 2x \cos 2x} dx = \int \frac{\sin x}{4 \sin x \cos x (1 - 2 \sin^2 x)} dx = \frac{1}{4} \int \frac{dx}{\cos x (1 - 2 \sin^2 x)}$.
Multiply numerator and denominator by $\cos x$: $I = \frac{1}{4} \int \frac{\cos x dx}{\cos^2 x (1 - 2 \sin^2 x)} = \frac{1}{4} \int \frac{\cos x dx}{(1 - \sin^2 x)(1 - 2 \sin^2 x)}$.
Let $\sin x = t$, then $\cos x dx = dt$. So $I = \frac{1}{4} \int \frac{dt}{(1 - t^2)(1 - 2t^2)}$.
Using partial fractions: $\frac{1}{(1 - t^2)(1 - 2t^2)} = \frac{A}{1 - t^2} + \frac{B}{1 - 2t^2}$. Solving gives $A = -1$ and $B = 2$.
$I = \frac{1}{4} \int \left( \frac{2}{1 - 2t^2} - \frac{1}{1 - t^2} \right) dt = \frac{1}{2} \int \frac{dt}{1 - (\sqrt{2}t)^2} - \frac{1}{4} \int \frac{dt}{1 - t^2}$.
Using $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right|$ and $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right|$, we get:
$I = \frac{1}{2} \cdot \frac{1}{2\sqrt{2}} \log \left| \frac{1 + \sqrt{2}t}{1 - \sqrt{2}t} \right| - \frac{1}{4} \cdot \frac{1}{2} \log \left| \frac{1+t}{1-t} \right| + c$.
Since $\frac{1}{2} \log \left| \frac{1+\sin x}{1-\sin x} \right| = \log |\sec x + \tan x|$, the result is $\frac{1}{4\sqrt{2}} \log \left| \frac{1 + \sqrt{2} \sin x}{1 - \sqrt{2} \sin x} \right| - \frac{1}{8} \log \left| \frac{1+\sin x}{1-\sin x} \right| + c = \frac{1}{4\sqrt{2}} \log \left| \frac{1 + \sqrt{2} \sin x}{1 - \sqrt{2} \sin x} \right| - \frac{1}{4} \log |\sec x + \tan x| + c$.

(B) Multiply the numerator and denominator by $e^x$:
$I = \int \frac{(x + 1)e^x}{xe^x(1 + xe^x)^2} dx$
Let $u = xe^x$, then $du = (e^x + xe^x) dx = e^x(1 + x) dx$.
Substituting these into the integral:
$I = \int \frac{du}{u(1 + u)^2}$
Using partial fractions: $\frac{1}{u(1 + u)^2} = \frac{A}{u} + \frac{B}{1 + u} + \frac{C}{(1 + u)^2}$
$1 = A(1 + u)^2 + Bu(1 + u) + Cu$
For $u = 0$, $A = 1$. For $u = -1$, $C = -1$. Comparing coefficients of $u^2$: $A + B = 0 \implies B = -1$.
$I = \int \left( \frac{1}{u} - \frac{1}{1 + u} - \frac{1}{(1 + u)^2} \right) du$
$I = \log|u| - \log|1 + u| + \frac{1}{1 + u} + c$
$I = \log \left| \frac{u}{1 + u} \right| + \frac{1}{1 + u} + c$
Substituting $u = xe^x$:
$I = \log \left( \frac{xe^x}{1 + xe^x} \right) + \frac{1}{1 + xe^x} + c$

(D) Given $I = \int \frac{\sin x}{2 \sin 2x \cos 2x} dx = \int \frac{\sin x}{4 \sin x \cos x (1 - 2 \sin^2 x)} dx = \frac{1}{4} \int \frac{dx}{\cos x (1 - 2 \sin^2 x)}$.
Multiply numerator and denominator by $\cos x$: $I = \frac{1}{4} \int \frac{\cos x dx}{\cos^2 x (1 - 2 \sin^2 x)} = \frac{1}{4} \int \frac{\cos x dx}{(1 - \sin^2 x)(1 - 2 \sin^2 x)}$.
Let $u = \sin x$, then $du = \cos x dx$. $I = \frac{1}{4} \int \frac{du}{(1 - u^2)(1 - 2u^2)}$.
Using partial fractions: $\frac{1}{(1 - u^2)(1 - 2u^2)} = \frac{A}{1 - u^2} + \frac{B}{1 - 2u^2}$.
$1 = A(1 - 2u^2) + B(1 - u^2)$. For $u^2 = 1$, $A = -1$. For $u^2 = 1/2$, $B = 2$.
$I = \frac{1}{4} \int \left( \frac{-1}{1 - u^2} + \frac{2}{1 - 2u^2} \right) du = -\frac{1}{4} \int \frac{du}{1 - u^2} + \frac{2}{4} \int \frac{du}{1 - (\sqrt{2}u)^2}$.
$I = -\frac{1}{8} \log \left| \frac{1+u}{1-u} \right| + \frac{1}{2} \cdot \frac{1}{2\sqrt{2}} \log \left| \frac{1+\sqrt{2}u}{1-\sqrt{2}u} \right| + c$.
Comparing with given form: $\alpha = -1/8$, $\beta = 1/(4\sqrt{2})$.
$\beta^2 = 1/32$. Thus, $32(\alpha + \beta^2) = 32(-1/8 + 1/32) = 32(-4/32 + 1/32) = 32(-3/32) = -3$.

(B) Let $I = \int \frac{dx}{\sin^2 x + \tan^2 x}$.
Since $\sin^2 x = \frac{\tan^2 x}{1 + \tan^2 x}$, we have $I = \int \frac{dx}{\frac{\tan^2 x}{1 + \tan^2 x} + \tan^2 x} = \int \frac{(1 + \tan^2 x) dx}{\tan^2 x + \tan^2 x(1 + \tan^2 x)} = \int \frac{\sec^2 x \ dx}{\tan^2 x + \tan^2 x + \tan^4 x} = \int \frac{\sec^2 x \ dx}{2\tan^2 x + \tan^4 x}$.
Let $u = \tan x$, then $du = \sec^2 x \ dx$. The integral becomes $I = \int \frac{du}{u^4 + 2u^2} = \int \frac{du}{u^2(u^2 + 2)}$.
Using partial fractions: $\frac{1}{u^2(u^2 + 2)} = \frac{A}{u^2} + \frac{B}{u^2 + 2}$. Solving gives $A = \frac{1}{2}$ and $B = -\frac{1}{2}$.
So, $I = \frac{1}{2} \int \frac{du}{u^2} - \frac{1}{2} \int \frac{du}{u^2 + 2} = \frac{1}{2} (-\frac{1}{u}) - \frac{1}{2} \cdot \frac{1}{\sqrt{2}} \tan^{-1} \left(\frac{u}{\sqrt{2}}\right) + c$.
Substituting $u = \tan x$, we get $I = -\frac{1}{2 \tan x} - \frac{1}{2\sqrt{2}} \tan^{-1} \left(\frac{\tan x}{\sqrt{2}}\right) + c$.