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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
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