Compute $\int \sqrt{1+4x^2} \, dx$ with Euler substitution In this post:
Computing $\int \sqrt{1+4x^2} \, dx$
someone mentioned Euler substitution to compute the following integral:
$$\int \sqrt{1+4x^2} \, dx$$
I tried to follow this advice and got very nice result, namely I substituted $\sqrt{1+4x^2}=t-2x$ which after raising to the square and reducing gives $x=\frac{t^2-1}{4t}$ and $t-2x=\frac{t^2+1}{2t}$, then derivative is equal to $\frac{dx}{dt}=\frac{t^2+1}{4t^2}$ and the whole integral:
$$\int \sqrt{1+4x^2} \, dx = \int \frac{(t^2+1)^2}{8t^3} \, dt$$
Could you please check my solution, because it seems a lot easier than all these trigonometric substitution (too easy which is suspicious...).
Thanks in advance.
 A: If you set $\sqrt{1+4x^2}=t-2x$, you have
$$
1+4x^2=t^2-4tx+4x^2
$$
so $4tx=t^2-1$ and therefore
$$
x=\frac{t^2-1}{4t}=\frac{t}{4}-\frac{1}{4t}
$$
Thus
$$
dx=\left(\frac{1}{4}+\frac{1}{4t^2}\right)\,dt=\frac{t^2+1}{4t^2}\,dt
$$
and
$$
\sqrt{1+4x^2}=t-\frac{t}{2}+\frac{1}{2t}=\frac{t^2+1}{2t}
$$
so the integral becomes
$$
\int\frac{(t^2+1)^2}{8t^3}\,dt=
\frac{1}{8}\int\left(t+\frac{2}{t}+\frac{1}{t^3}\right)\,dt
$$
that's elementary.
Yes, your computation is right.

The alternative way is setting $2x=\sinh(t/2)$, so
$$
\sqrt{1+4x^2}=\cosh\frac{t}{2},
\qquad
dx=\frac{1}{4}\cosh\frac{t}{2}\,dt
$$
and the integral is
$$
\frac{1}{4}\int\cosh^2\frac{t}{2}\,dt=
\frac{1}{8}\int(\cosh t-1)\,dt
$$
remembering that $2\cosh^2\frac{t}{2}+1=\cosh t$.

Another (tricky) way is to do $2x=t$, that reduces to computing, up to a scalar factor,
$$\DeclareMathOperator{\arsinh}{arsinh}
I=\int\sqrt{1+t^2}\,dt=\int\frac{1+t^2}{\sqrt{1+t^2}}\,dt=
\arsinh t+\int t\frac{t}{\sqrt{1+t^2}}\,dt=
\arsinh t+t\sqrt{1+t^2}-I
$$
(the last one by parts).
A: Hint
$$x=\frac{1}{2}\tan\theta$$
$$I=\frac12\int\sec^3\theta\,d\theta $$
