What is #int (4x ) / sqrt(4-x^4) dx#?

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Answer 1 · 2018-05-28T19:21:04

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

Here,

#I=int(4x)/sqrt(4-x^4)dx#

#x^2=2sinu=>2xdx=2cosudu=>4xdx=4cosudu#

#andsinu=x^2/2=>color(blue)(u=sin^-1(x^2/2)#

So,

#I=int(4cosu)/sqrt(4-4sin^2u)du#

#=int(4cosu)/sqrt(4cos^2u)du#

#=int(4cosu)/(2cosu)du#

#=int2du#

#=2u+c...to where,color(blue)(u=sin^-1(x^2/2)#

Hence,

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