How to integrate the following ??

Question

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

966 views around the world

You can reuse this answer
Creative Commons License

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#