`int sqrt(1-x^2)/x^4 dx` Find the indefinite integral

Given
,

`int sqrt(1-x^2)/x^4 dx`

By applying Integration by
parts we can solve the given integral

so,

let `u=
sqrt(1-x^2)` `,v' = (1/x^4) `

=>` u' = (sqrt(1-x^2) )'`

=> =`d/dx (sqrt(1-x^2)) `

let `t=1-x^2 `

so,

`d/dx (sqrt(1-x^2))`

=`d/dx
(sqrt(t))`

= `d/(dt) sqrt(t) * d/dx (t)`        [as `d/dx f(t) = d/(dt) f(t)
(dt)/dx` ]

=  `[(1/2)t^((1/2)-1) ]*(d/dx (1-x^2))`

= 
`[(1/2)t^(-1/2)]*(-2x)`

=`[1/(2sqrt(1-x^2 ))]*(-2x)`

=`-x/sqrt(1-x^2)`

so,  `u' = -x/sqrt(1-x^2)` and as ` v'=(1/x^4)`
so

`v = int 1/x^4 dx `

= `int x^(-4) dx`

= `(x^(-4+1))/(-4+1) `

=`(x^(-3))/(-3)`

=
`-(1/(3x^3))`

 

so , let us see the values
altogether.

`u= sqrt(1-x^2) ,u' = -x/sqrt(1-x^2)` and` v' = (1/x^4)
,v=-(1/(3x^3))`

 

Now ,by applying the integration by
parts `int uv' ` is given as

 `int uv' = uv - int u'v`

then,

`int sqrt(1-x^2)/x^4 dx `

= `(sqrt(1-x^2))
(-(1/(3x^3))) - int (-x/sqrt(1-x^2))(-(1/(3x^3))) dx `

= `(sqrt(1-x^2))
(-(1/(3x^3))) -(-...