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

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Given ,

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

By applying Integration by parts we can solve the given integral


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 `


`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)`


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) `


= `-(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`


`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))) -(-...

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