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#