How to integrate the following ??

Question

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

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Answer 1 · 2017-10-01T16:09:34

The answer is $=1/2arcsin((2x^2+1)/sqrt5)+C$

Perform this integral by substitution

Let $x=u^2$ , $=>$ , $dx=2udu$

$sqrt(-x^4-x^2+1)= sqrt(-u^2-u+1)$

$int(2xdx)/sqrt(-x^4-x^2+1)=1/2int(du)/sqrt(-u^2-u+1)$

$-u^2-u+1=1-u-u^2=1-(u^2+u)=(1+1/4-(u^2+u+1/4))$

$=(5/4-(u+1/2)^2)$

So,

$1/2int(du)/sqrt(-u^2-u+1)=1/2int(du)/sqrt(5/4-(u+1/2)^2)$

Let $v=(2u+1)/sqrt5$ , $=>$ , $dv=(2du)/sqrt5$

$5/4-(u+1/2)^2=5/4-5v^2=5/4(1-v^2)$

$1/2int(du)/sqrt(5/4-(u+1/2)^2)=1/2int(dv)/sqrt(1-v^2)$

This is a standard integral

$1/2int(dv)/sqrt(1-v^2)=1/2arcsinv$

$=1/2sin(2u+1)/sqrt5$

$=1/2arcsin((2x^2+1)/sqrt5)+C$