Question

Obtain: $\int \frac{(3x+2)}{(x+1)(x+2)(x-3)}dx$

Answer 1

Let $\frac{(3x+2)}{(x+1)(x+2)(x-3)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x-3}$

$\frac{(3x+2)}{(x+1)(x+2)(x-3)}=\frac{A(x^2-x-6)+B(x^2-2x-3)+C(x^2+3x+2)}{(x+1)(x+2)(x-3)}$

$\frac{(3x+2)}{(x+1)(x+2)(x-3)}=\frac{x^2(A+B+C)+x(-A-2B+3C)+(-6A-3B+2C)}{(x+1)(x+2)(x-3)}$

$\therefore A+B+C=0$

$\therefore A+2B-3C=-3$

$\therefore 6A+3B-2C=-2$

Solving these eq, we get:

$A=\frac{1}{4}$

$B=-\frac{4}{5}$

$C=\frac{11}{20}$

$\therefore I=\int \frac{A}{x+1}dx+\int \frac{B}{x+2}dx+\int \frac{C}{x-3}dx$

$\therefore I=\frac{1}{4}ln(x+1)-\frac{4}{5}ln(x+2)+\frac{11}{20}ln(x-3)+c$ $\left(Ans\right)$