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int e^(tan^-1x)(1+x+x^2) d(cot^-1x) is e...

`int e^(tan^-1x)(1+x+x^2) d(cot^-1x)` is equal to

A

` - e^(tan^(-1)x)+c`

B

` e^(tan^(-1)x)+c`

C

`-x e^(tan^(-1)x)+c`

D

`x e^(tan^(-1)x)+c`

Text Solution

Verified by Experts

The correct Answer is:
C
`I=int e^(tan^(-1)x) (1+x+x^(2))(-((1)/(1+x^(2)))dx)`
`= -int e^(tan^(-1)x)(1+(x)/(1+x^(2)))dx`
`= -int e^(tan^(-1)x)dx-int x(e^(tan^(-1)x))/(1+x^(2))dx`
`= -int e^(tan^(-1)x)dx-x e^(tan^(-1)x)+int e^(tan^(-1)x) dx+C`
`= -x e^(tan^(-1)x)+C`
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