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int(dx)/(x^(2)sqrt(16-x^(2))) has the va...

`int(dx)/(x^(2)sqrt(16-x^(2)))` has the value equal to

A

`C-(1)/(4)tan^(-1)sec((x)/(4))`

B

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

C

`C-(sqrt(16-x^(2)))/(16x)`

D

`(sqrt(16-x^(2)))/(16x)+C`

Text Solution

Verified by Experts

The correct Answer is:
C
`I=int(1)/(x^(2)sqrt(16-x^(2)))dx`
Put `x=(1)/(t),`
`dx=-(1)/(t^(2))dt therefore I=int(-(1)/(t^(2))dt)/((1)/(t)xx(1)/(t^(2))sqrt(16t^(2)-1))=int(-tdt)/(sqrt(16t^(2)-1))`
Let `16t^(2)-1=u^(2), 32tdt = 2u du,`
`tdt=(u)/(16)du thereforeI=-(1)/(16)int(udu)/(u)=-(u)/(16)+C=-(sqrt(16-x^(2)))/(16x)+C`
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