How do you evaluate the integral of $int x/(1-x^4)^(1/2) dx$ ?

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Answer 1 · 2018-06-17T19:45:29

$I=1/2sin^-1(x^2)+c$

Here,

$I=intx/sqrt(1-x^4)dx=intx/sqrt(1-(x^2)^2)dx$

Subst . $x^2=u=>2xdx=du=>xdx=1/2du$

So,

$I=1/2int1/sqrt(1-u^2)du$

$=1/2sin^-1u+c ,where, u=x^2$

$=1/2sin^-1(x^2)+c$

How do you evaluate the integral of int x/(1-x^4)^(1/2) dxint x/(1-x^4)^(1/2) dx?