Question

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

Answer 1

$I=\int \frac{x^2+x+3}{(x-2)(x+1)}dx$

Since, the degree of numerator is equal to the degree of denominator

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

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

$\Rightarrow x^2+x+3=(x-2)(x+1)+A(x+1)+B(x-2)$ ....(2)

Put $x=2$ in eqn (2),

$\Rightarrow A=3$

Put $x=-1$ in eqn (2),

$\Rightarrow B=-1$

Put these values in eqn (1),

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

Integrating both sides , we get

$I=-\int \frac{1}{x+1}+3\int \frac{1}{x-2}+\int dx=-\log (x+1)+3\log (x-2)+x+C$