Question

Evaluate $\int \frac{3x^2+4x+3}{(x-2)(x^2+3x+2)}dx$.

Answer 1

Let $I=\int \frac{3x^2+4x+3}{(x-2)(x^2+3x+2)}dx$

Now,
$\frac{3x^2+4x+3}{(x-2)(x^2+3x+2)}=\frac{3x^2+4x+3}{(x-2)(x+1)(x+2)}$

By using partial fraction,
$\frac{3x^2+4x+3}{(x-2)(x+1)(x+2)}=\frac{A}{x-2}+\frac{B}{x+1}+\frac{C}{x+2}$
$\Rightarrow 3x^2+4x+3=A(x^2+3x+2)+B(x^2-4)+C(x^2-x-2)$

On comparing coeficients of constant, $x$ and $x^2$, we get:
$A+B+C=3...\left(1\right)3A-C=4...\left(2\right)2A-4A-2C=3...\left(3\right)$

By solving the equations (1),(2) and (3), we get:
$A=\frac{23}{12},B=-\frac{2}{3},C=\frac{7}{4}$

$\therefore I=\frac{23}{12}\int \frac{1}{x-2}dx-\frac{2}{3}\int \frac{1}{x+1}dx+\frac{7}{4}\int \frac{1}{x+2}dx$
$=\frac{23}{12}{\log }_e(x-2)-\frac{2}{3}{\log }_e(x+1)+\frac{7}{4}{\log }_e(x+2)+C$