If $\int \frac{1}{(x+1)(x-2)} d x=A \log

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Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Indefinite Integration

Question:

If $\int \frac{1}{(x+1)(x-2)} d x=A \log _e(x+1)+B \log _e(x-2)+C$, then

Options:

$A+B=0$

$A-B=0$

$A B=1$

$A B=-1$

Correct Answer:

$A+B=0$

Explanation:

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 $B=\frac{1}{3} \Rightarrow A+B=0$