Question

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

• A. $A+B=0$
• B. $AB=-1$
• C. $A:B=-1$
• D. none of these

Answer 1

The correct option is A $A+B=0$

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

This can be written as

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

$\Rightarrow \frac{-1}{3}\int \frac{1}{x+1}dx+\frac{1}{3}\int \frac{1}{x-2}dx$

$\Rightarrow \frac{-1}{3}\log (x+1)+\frac{1}{3}\log (x-2)+C$

Thus comparing this whith given expression,

$\Rightarrow A=\frac{-1}{3},B=\frac{1}{3}$

$\Rightarrow A+B=0$