# 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

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))) -(-...

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