How do you integrate #int x /sqrt( 16+x^4 )dx# using trigonometric substitution?

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Answer 1 · 2018-05-19T16:44:08

Use the substitution #x^2=4tantheta#.

Let

#I=intx/sqrt(16+x^4)dx#

Apply the substitution #x^2=4tantheta#:

#I=1/2intd theta#

The integral is trivial:

#I=1/2theta+C#

Reverse the substitution:

#I=1/2tan^(-1)(x^2/4)+C#