<s>[INST] Computing $\int \sqrt{1+4x^2} \, dx$ It was said that we did not learn how to take this integral in class yet, and that we should just use a graphing utility to find the integral, so being me i took that as a challenge and tried to take it using methods i knew.
$$
\int \sqrt{1+4x^2} dx
$$
$$
\sqrt{1+4x^2} = \sqrt{(2x+1)(2x-1)+2}
$$
$u = 2x-1, du = \frac{1}{2} dx$
$$
\int \sqrt{u(u+2) + 2} du
$$
$$
\int \sqrt{(u+1)^2 + 1} du
$$
Nothing there
$$
\int\sqrt{1+4x^2}dx = x\sqrt{1+4x^2} - \int \frac{4x^2}{\sqrt{1+4x^2}} dx
$$
Tried for a while to put the other integral into a workable form and got nowhere.
So what method is used to take this integral?
 [/INST] Substitute $x = \frac 12 \sinh \phi$ to get
$$\int \sqrt{1+4x^2}\mathrm dx = \int \cosh^2 \phi\ \mathrm d\phi = \int \left(\frac 12 \cosh 2\phi + \frac 12\right)\mathrm d\phi = \frac 18\sinh 2\phi + \frac 12\phi + C = \boxed{\displaystyle\frac x2\sqrt{1+4x^2} + \operatorname{arcsinh}2x + C}$$
Note the similarity between this integral and
$$\int \sqrt{1-4x^2} = \frac x2\sqrt{1-4x^2} + \arcsin 2x + C$$
If you didn't remember the identity $\cosh^2 x = \frac 12\cosh 2x + \frac 12$ you could have used its definition:
$$\cosh^2 x = \frac {\left(e^x + e^{-x}\right)^2} 4$$
It's an easy integral then, but the final substitution is indeed ugly and does not work out as nicely as above.
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