Can anybody help with the integral: $\int \frac{\sqrt{1-x^2}}{\sqrt{1+x^2}} dx$ please? I have searched online. I have got a solution from Wolfram but I don't understand it, or how to reach it. If anyone has a method that would be fantastic, thanks!

This is an elliptic integral, but it is interesting to point out what is the explicit value of the integral over $(0,1)$.
$$ \int_{0}^{1}\sqrt{\frac{1-x^2}{1+x^2}}\,dx = \frac{1}{2}\int_{0}^{1}\sqrt{\frac{1-x}{x(1+x)}}\,dx=\frac{1}{2}\int_{0}^{1}\sqrt{\frac{x}{(1-x)(2-x)}}\,dx$$
equals
$$ \int_{0}^{1}\frac{x^2}{\sqrt{(1-x^2)(2-x^2)}}\,dx =\int_{0}^{\pi/2}\frac{\sin^2\theta}{\sqrt{2-\sin^2\theta}}\,d\theta$$
or
$$2\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{2-\sin^2\theta}}-\int_{0}^{\pi/2}\sqrt{2-\sin^2\theta}\,d\theta=\sqrt{2}\left[K\left(\tfrac{1}{2}\right)-E\left(\tfrac{1}{2}\right)\right]$$
by representing the complete elliptic integrals according to Mathematica's notation, such that the argument is the elliptic modulus. A series representation for the RHS is given by
$$ \frac{\pi}{\sqrt{2}}\sum_{n\geq 1}\left[\frac{1}{4^n}\binom{2n}{n}\right]^2\frac{2n}{1-2n}\cdot\frac{1}{2^n}$$
which due to $\frac{\binom{2n}{n}}{4^n}\approx\frac{1}{\sqrt{\pi n}}$ for large values of $n$, can be approximated by
$$ \frac{\pi}{4\sqrt{2}}+\sqrt{2}\sum_{n\geq 2}\frac{n}{2^n(2n-1)}=\frac{\pi-4}{4\sqrt{2}}+\log(1+\sqrt{2}).$$
The Cauchy-Schwarz inequality provides a simpler (and tighter) upper bound, namely
$$ \int_{0}^{1}\sqrt{\frac{1-x^2}{1+x^2}}\,dx<\sqrt{\int_{0}^{1}(1-x^2)\,dx\int_{0}^{1}\frac{dx}{1+x^2}}=\sqrt{\frac{\pi}{6}}.$$
On the other hand both $K\left(\tfrac{1}{2}\right)$ and $E\left(\tfrac{1}{2}\right)$ can be computed through the $\text{AGM}$ and are explicitly related to the $\Gamma$ function. Indeed the former is given by a multiple of $B\left(\tfrac{1}{4},\tfrac{1}{4}\right)$ and the latter is related to the former via Legendre's relation. This gives
$$ \int_{0}^{1}\sqrt{\frac{1-x^2}{1+x^2}}\,dx = \color{blue}{\frac{\Gamma\left(\tfrac{1}{4}\right)^2}{4\sqrt{2\pi}}-\frac{\pi\sqrt{2\pi}}{\Gamma\left(\tfrac{1}{4}\right)^2}}=\color{red}{\frac{\pi}{2\,\text{AGM}(1,\sqrt{2})}-\frac{\text{AGM}(1,\sqrt{2})}{2}}$$
which can be proved through Fourier-Legendre expansions, too.
