If $\int \frac{1}{(x+1)(x-2)} d x=A \log _e(x+1)+B \log _e(x-2)+C$, then
Answer & explanation
Correct answer: option 1
Let,
$I=\int \frac{1}{(x+1)(x-2)} d x=\int \frac{1}{3}\left(\frac{1}{x-2}-\frac{1}{x+1}\right) d x$
$\Rightarrow I=\frac{1}{3} \log _e(x-2)-\frac{1}{3} \log _e(x+1)+C$
∴ $A=-\frac{1}{3}$ and $B=\frac{1}{3} \Rightarrow A+B=0$