Question

$\int \frac{dx}{(x+1)(x-2)}=A\log (x+1)+B\log (x-2)+C$, where

• A. $A+B=0$
• B. $AB=0$
• C. $\frac{A}{B}=-1$
• D. none of these

Answer 1

Answer: (A), (C)

$A+B=0$
$\frac{A}{B}=-1$

Given, $\int \frac{dx}{(x+1)(x-2)}=A\log (x+1)+B\log (x-2)+C$ ...(1)
$\frac{1}{(x+1)(x-2)}=\frac{C}{(x+1)}+\frac{D}{(x-2)}$ ....(2)
$\Rightarrow 1=C(x-2)+D(x+1)$
When $x=2,D=\frac{1}{3}$
When $x=-1,C=-\frac{1}{3}$
Put these values in (2)
$\frac{1}{(x+1)(x-2)}=\frac{-1}{3(x+1)}+\frac{1}{3(x-2)}$
$\int \frac{dx}{(x+1)(x-2)}=\frac{-1}{3}\log (x+1)+\frac{1}{3}\log (x-2)$
On comparing with (1), we get
$A=-\frac{1}{3}$
and $B=\frac{1}{3}$
$\Rightarrow A+B=0$
and $\frac{A}{B}=-1$