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Evaluate $\int \sqrt{1+x^2} d x$.
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Let $x=\tan u, d x=\sec ^2 u d u$, so
$\int \sqrt{1+x^2} d x=\int \sqrt{1+\tan ^2 u} \sec ^2 u d u=\int \sqrt{\sec ^2 u} \sec ^2 u d u .$
Since $u=\arctan (x),-\pi / 2 \leq u \leq \pi / 2$ and $\sec u \geq 0$, so $\sqrt{\sec ^2 u}=\sec u$. Then
$\int \sqrt{\sec ^2 u} \sec ^2 u d u=\int \sec ^3 u d u .$
In problems of this type, two integrals come up frequently: $\int \sec ^3 u d u$ and $\int \sec u d u$. Both have relatively nice expressions but they are a bit tricky to discover.

First we do $\int \sec u d u$, which we will need to compute $\int \sec ^3 u d u$ :
\begin{aligned} \int \sec u d u &=\int \sec u \frac{\sec u+\tan u}{\sec u+\tan u} d u \\ &=\int \frac{\sec ^2 u+\sec u \tan u}{\sec u+\tan u} d u . \end{aligned}
Now let $u=\sec u+\tan u$, dw $=\sec u \tan u+\sec ^2 u d u$, exactly the numerator of the function we are integrating. Thus
\begin{aligned} \int \sec u d u=\int \frac{\sec ^2 u+\sec u \tan u}{\sec u+\tan u} d u &=\int \frac{1}{w} d u=\ln |w|+C \\ &=\ln |\sec u+\tan u|+C . \end{aligned}
Now for $\int \sec ^3 u d u:$
\begin{aligned} \sec ^3 u &=\frac{\sec ^3 u}{2}+\frac{\sec ^3 u}{2}=\frac{\sec ^3 u}{2}+\frac{\left(\tan ^2 u+1\right) \sec u}{2} \\ &=\frac{\sec ^3 u}{2}+\frac{\sec u \tan ^2 u}{2}+\frac{\sec u}{2}=\frac{\sec ^3 u+\sec u \tan ^2 u}{2}+\frac{\sec u}{2} . \end{aligned}
We already know how to integrate sec: $u$, so we just need the first quotient. This is "simply" a matter of reognizing the product rule in action:
$\int \sec ^3 u+\sec u \tan ^2 u d u=\sec u \tan u$
So putting these together we get
$\int \sec ^3 u d u=\frac{\sec u \tan u}{2}+\frac{\ln |\sec u+\tan u|}{2}+C_?$
and reverting to the original variable $x$ :
\begin{aligned} \int \sqrt{1+x^2} d x &=\frac{\sec u \tan u}{2}+\frac{\ln |\sec u+\tan u|}{2}+C \\ &=\frac{\sec (\arctan x) \tan (\arctan x)}{2}+\frac{\ln |\sec (\arctan x)+\tan (\arctan x)|}{2}+C \\ &=\frac{x \sqrt{1+x^2}}{2}+\frac{\ln \left|\sqrt{1+x^2}+x\right|}{2}+C, \\ \text { using } \tan (\arctan x) &=x \text { and } \sec (\arctan x)=\sqrt{1+\tan ^2(\arctan x)}=\sqrt{1+x^2} . \end{aligned}
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