Im stuck at integrating algebraic substitution

Our function is:

\(\int \frac{x^2}{\left(x^2+1\right)^{\frac{3}{2}}}dx\)

I looked at this for 20 minutes, and realized that algebraic substitution is NOT the right way to do it.

Instead, you have to perform trigonometric substitution:

\(x=\tan \left(u\right)\)

Solve for \(u\), then find derivative with respect to x:

\(\tan ^{-1}x=u\)

\(\frac{du}{dx}=\sec ^2\left(u\right)\)

Solve for dx:

\(du=\sec ^2\left(u\right)\cdot dx\)

Now your expression is:

\(\int \frac{\tan ^2\left(u\right)}{\left(\tan ^2u+1\right)^{\frac{3}{2}}}\sec ^2\left(u\right)du\)

At the bottom:

Simplify using \(\tan ^2(u)+1=\sec ^2(u)\)

We get:

\(\int \frac{\tan ^2\left(u\right)}{\left[\sec ^2\left(u\right)\right]^{\frac{3}{2}}}\sec ^2\left(u\right)du\)

\(\int \frac{\tan ^2\left(u\right)}{\sec ^3\left(u\right)}\sec ^2\left(u\right)du\)

\(\int \frac{\tan ^2\left(u\right)}{\sec \left(u\right)}du\)

Same as:

\(\int \cos \left(u\right)\tan ^2\left(u\right)du\)

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Please let me know if you're following