Solutions - 0050-05

(1) Notice $x^2+1$ pattern, so we should make use of the identity ${\tan }^2\theta +1={\sec }^2\theta$.

Select $x=\tan \theta$ and thus $dx={\sec }^2\theta d\theta$. Then:

$${(x^2+1)}^{3/2}\gg \gg {({\sec }^2\theta )}^{3/2}\gg \gg {\sec }^3\theta$$

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(2) Convert to $\theta$ and integrate:

$$\begin{array}{c}\int \frac{x^2}{{(x^2+1)}^{3/2}}dx\gg \gg \int \frac{{\tan }^2\theta {\sec }^2\theta }{{\sec }^3\theta }d\theta \\ \\ \gg \gg \int \frac{{\tan }^2\theta }{\sec \theta }d\theta \gg \gg \int \frac{{\sec }^2\theta -1}{\sec \theta }d\theta \\ \\ \gg \gg \int \sec \theta -\cos \theta d\theta \gg \gg \ln |\tan \theta +\sec \theta |-\sin \theta +C\end{array}$$

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(3) Convert back to terms of $x$:

Draw a triangle expressing $\tan \theta ={\displaystyle \frac{x}{1}}$:

Therefore $\sec \theta =\sqrt{x^2+1}$ and $\sin \theta ={\displaystyle \frac{x}{\sqrt{x^2+1}}}$. Then:

$$\begin{array}{c}\ln |\tan \theta +\sec \theta |-\sin \theta +C\\ \\ \gg \gg \ln |x+\sqrt{x^2+1}|-\frac{x}{\sqrt{x^2+1}}+C\end{array}$$