TITLE: Integral of $\frac{2x}{\sqrt{1+x^2}}$
QUESTION [5 upvotes]: I'm trying to find the following indefinite integral:
$$
\int \frac{2x dx}{\sqrt{1+x^2}}
$$
I have already done it via u-substitution, however, I've been asked to solve it with two methods.
I think I can do it by trig substitution as well, however I'm not sure how to finish.
I know I can substitute $x$ in the square root for $\tan\theta$ by the trig substitution rules, but I'm still a bit confused by the $2x$ in the numerator. Does that go to $2\tan\theta$? And how would I finish this?

REPLY [1 votes]: Put $1+x^2=z^2$ and hence $xdx=zdz$ to get $$\int\frac{2x}{\sqrt{1+x^2}}dx=\int\frac{2z}{z}dz=2z+c=2\sqrt{1+x^2}+c$$
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