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,

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

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