# How do you integrate #int x^2 /sqrt( 16+x^4 )dx# using trigonometric substitution?

This cannot be integrated using elementary functions.

We have:

The more we continue, we see that this cannot be integrated using elementary functions.

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To integrate ( \int \frac{x^2}{\sqrt{16+x^4}} , dx ) using trigonometric substitution:

- Recognize that the expression under the square root resembles a sum of squares, so we can use trigonometric substitution.
- Let ( x^2 = 4 \tan(\theta) ), which implies ( dx = 2 \tan(\theta) \sec^2(\theta) , d\theta ).
- Substitute ( x^2 = 4 \tan(\theta) ) and ( dx = 2 \tan(\theta) \sec^2(\theta) , d\theta ) into the integral.
- The integral becomes ( \int \frac{4\tan^2(\theta) \sec^2(\theta)}{\sqrt{16 + 4\tan^2(\theta)}} , d\theta ).
- Simplify to get ( \int \frac{4\tan^2(\theta) \sec^2(\theta)}{\sqrt{4(4 + \tan^2(\theta))}} , d\theta ).
- Simplify further to ( \int \frac{4\tan^2(\theta) \sec^2(\theta)}{2\sqrt{4 + \tan^2(\theta)}} , d\theta ).
- Reduce to ( \int 2 \tan^2(\theta) , d\theta ).
- Use trigonometric identity ( \tan^2(\theta) = \sec^2(\theta) - 1 ) to rewrite the integral as ( \int 2(\sec^2(\theta) - 1) , d\theta ).
- Integrate term by term to get ( 2 \tan(\theta) - 2\theta + C ).
- Substitute back ( \theta = \arctan\left(\frac{x^2}{4}\right) ) to get the final result ( \boxed{2 \tan\left(\arctan\left(\frac{x^2}{4}\right)\right) - 2\arctan\left(\frac{x^2}{4}\right) + C} ).

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