Q.Find $\int \dfrac{3x - 2}{(x+1)^2 (x+3)}\, dx$

Reduce the improper rational integrand by division then partial fractions; the OR integral evaluates to $\pi$ using the King's-rule symmetry $x\to\pi-x$.

Main question: $x^2-5x+6=(x-2)(x-3)$. Since the numerator and denominator have the same degree, divide first:

$\frac{x^2+1}{x^2-5x+6}=1+\frac{(x^2+1)-(x^2-5x+6)}{x^2-5x+6}=1+\frac{5x-5}{(x-2)(x-3)}.$

Partial fractions: $\dfrac{5x-5}{(x-2)(x-3)}=\dfrac{A}{x-2}+\dfrac{B}{x-3}$, so $5x-5=A(x-3)+B(x-2)$.

• $x=2$: $5=-A\Rightarrow A=-5$
• $x=3$: $10=B\Rightarrow B=10$

So the integrand is $1-\dfrac{5}{x-2}+\dfrac{10}{x-3}$, and

$\int\frac{x^2+1}{x^2-5x+6}dx = x-5\ln|x-2|+10\ln|x-3|+C.$

OR: $I=\displaystyle\int_0^\pi\frac{x\,dx}{1+\sin x}$. Using $\int_0^a f(x)dx=\int_0^a f(a-x)dx$ with $\sin(\pi-x)=\sin x$:

$I=\int_0^\pi\frac{(\pi-x)\,dx}{1+\sin x}=\pi\int_0^\pi\frac{dx}{1+\sin x}-I\ \Rightarrow\ 2I=\pi\int_0^\pi\frac{dx}{1+\sin x}.$ …

Write the numerator as (derivative of denominator) minus a constant, split the integral, and use partial fractions for the remaining rational part.

(i) $\displaystyle\int \frac{2x}{x^2+3x+2}dx$:

Note $\dfrac{d}{dx}(x^2+3x+2) = 2x+3$. Write $2x = (2x+3) - 3$:

$\int \frac{2x}{x^2+3x+2}dx = \int\frac{2x+3}{x^2+3x+2}dx - 3\int\frac{dx}{x^2+3x+2}.$

The first integral is $\log|x^2+3x+2|$ directly (numerator is the derivative of the denominator).

For the second, factor $x^2+3x+2=(x+1)(x+2)$, and use partial fractions:

$\frac{1}{(x+1)(x+2)} = \frac{1}{x+1} - \frac{1}{x+2} \implies \int\frac{dx}{x^2+3x+2} = \log|x+1| - \log|x+2|.$

Combining:

$\int\frac{2x}{x^2+3x+2}dx = \log|x^2+3x+2| - 3\left[\log|x+1|-\log|x+2|\right] + C.$

Since $x^2+3x+2 = (x+1)(x+2)$, $\log|x^2+3x+2| = \log|x+1| + \log|x+2|$, so

$= \log|x+1|+\log|x+2| - 3\log|x+1| + 3\log|x+2| + C = -2\log|x+1| + 4\log|x+2| + C.$

…