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Evaluate \(\int \sqrt{1+x^2} d x\).
in Mathematics by Platinum (141,884 points) | 19 views

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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 \(x\) :
\[
\begin{aligned}
\int \sqrt{1+x^2} d x &=\frac{\sec u \tan u}{2}+\frac{\ln |\sec u+\tan u|}{2}+C \\
&=\frac{\sec (\arctan x) \tan (\arctan x)}{2}+\frac{\ln |\sec (\arctan x)+\tan (\arctan x)|}{2}+C \\
&=\frac{x \sqrt{1+x^2}}{2}+\frac{\ln \left|\sqrt{1+x^2}+x\right|}{2}+C, \\
\text { using } \tan (\arctan x) &=x \text { and } \sec (\arctan x)=\sqrt{1+\tan ^2(\arctan x)}=\sqrt{1+x^2} .
\end{aligned}
\]
by Platinum (141,884 points)

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