How do you integrate $\int \frac{x}{\sqrt{16 + x^{4}}} d x$ $int x /sqrt( 16+x^4 )dx$ using trigonometric substitution?

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Answer 1 · 2018-05-19T16:44:08

Use the substitution $x^{2} = 4 tan θ$ $x^2=4tantheta$ .

Let

$I = \int \frac{x}{\sqrt{16 + x^{4}}} d x$ $I=intx/sqrt(16+x^4)dx$

Apply the substitution $x^{2} = 4 tan θ$ $x^2=4tantheta$ :

$I = \frac{1}{2} \int d θ$ $I=1/2intd theta$

The integral is trivial:

$I = \frac{1}{2} θ + C$ $I=1/2theta+C$

Reverse the substitution:

$I = \frac{1}{2} {tan}^{- 1} (\frac{x^{2}}{4}) + C$ $I=1/2tan^(-1)(x^2/4)+C$

How do you integrate int x /sqrt( 16+x^4 )dx∫x√16+x4dxint x /sqrt( 16+x^4 )dx using trigonometric substitution?