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int (dx)/((x+1)(x-2))=A log (x+1)+B log ...

`int (dx)/((x+1)(x-2))=A log (x+1)+B log (x-2)+C`, where

A

A+B=0

B

A - B = 0

C

AB =1

D

AB =-1

Text Solution

Verified by Experts

The correct Answer is:
A
Let `I=int(1)/((x+1)(x-2))dx=int(1)/(3)((1)/(x-2)-(1)/(x+1))`dx
`rArrI=(1)/(3)log_(e)(x-2)-(1)/(3)log_(e)(x+1)+C`
`thereforeA=-(1)/(3)andB=(1)/(3)rArrA+B=0`
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OBJECTIVE RD SHARMA ENGLISH-INDEFINITE INTEGRALS-Solved Example
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