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

Options

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`

Answer

C

Explanation

`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`