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int(1)/(16x^(2)+9)dx is equal to...

`int(1)/(16x^(2)+9)dx` is equal to

A

`(1)/(3)tan^(-1)((4x)/(3))+C`

B

`(1)/(4)tan^(-1)((4x)/(3))+C`

C

`(1)/(12)tan^(-1)((4x)/(3))+C`

D

`(1)/(12)tan^(-1)((3x)/(4))+C`

Text Solution

Verified by Experts

The correct Answer is:
C
`int(1)/(16x^(2)+9)dx=(1)/(16)int(1)/(x^(2)+((3)/(4))^(2))dx`
`=(1)/(16)xx(4)/(3)tan^(-1)((x)/(3//4))+C=(1)/(12)tan^(-1)((4x)/(3))+C`
(c) `int[sin(logx)+cos(logx)]dx`
`=int(d)/(dx){x sin(logx)}dx=x sin (logx)+C`
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