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Just got out of a Calculus 2/ Lineal Algebra exam and im still wondering about the first question:

∫arcsin(2×

I used integration by parts letting u=arcsin(2×

=xarcsin(2×

this second integral is what I can´t solve.

I tried doing trigonometric substitution where sin∅=2x

If anyone out there has any clue, its much appreciated! :)

∫arcsin(2×

^{2})dxI used integration by parts letting u=arcsin(2×

^{2}) and so du=dx/sqrt[1-4x^{4}], v=x=xarcsin(2×

^{2}) - ∫[x/sqrt[1-4x^{4}]dxthis second integral is what I can´t solve.

I tried doing trigonometric substitution where sin∅=2x

^{2}, and then later used a trig identity to integrate... but I really think I messed up somewhere.If anyone out there has any clue, its much appreciated! :)

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