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!

If the integral is a definite integral with the interval of integration being from $0$ to $1$, here is an approach that makes use of the following form for the Beta function $\operatorname{B} (x,y)$:
$$\operatorname{B} (x,y) = \int_0^1 t^{x - 1} (1 - t)^{y - 1} \, dt, \qquad x,y > 0. \tag1$$
Let
$$I = \int_0^1 \sqrt{\frac{1 - x^2}{1 + x^2}} \, dx.$$
Enforcing a substitution of $x \mapsto \sqrt{x}$ yields
$$I = \frac{1}{2} \int_0^1 \sqrt{\frac{1 - x}{1 + x}} \, \frac{dx}{\sqrt{x}}.$$
Rearranging the integrand we have
\begin{align}
I &= \frac{1}{2} \int_0^1 \sqrt{\frac{1 - x}{1 + x}} \cdot \sqrt{\frac{1 - x}{1 - x}} \, \frac{dx}{\sqrt{x}}\\
&= \frac{1}{2} \int_0^1 \frac{1 - x}{\sqrt{1 - x^2}} \, \frac{dx}{\sqrt{x}}\\
&= \frac{1}{2} \int_0^1 \frac{dx}{\sqrt{1 - x^2} \sqrt{x}} - \frac{1}{2} \int_0^1 \frac{\sqrt{x}}{\sqrt{1 - x^2}} \, dx. \qquad \qquad \qquad \qquad (*)
\end{align}
Enforcing a substitution of $x \mapsto \sqrt{x}$ in each of the above two integrals leads to
\begin{align}
I &= \frac{1}{4} \int_0^1 x^{-3/4} (1 - x)^{-1/2} \, dx - \frac{1}{4} \int_0^1 x^{-1/4} (1 - x)^{-1/2} \, dx\\
&= \frac{1}{4} \int_0^1 x^{1/4 - 1} (1 - x)^{1/2 - 1} \, dx - \frac{1}{4} \int_0^1 x^{3/4 - 1} (1 - x)^{1/2 - 1} \, dx\\
&= \frac{1}{4} \operatorname{B} \left (\frac{1}{4}, \frac{1}{2} \right ) - \frac{1}{4} \operatorname{B} \left (\frac{3}{4}, \frac{1}{2} \right ) \tag2\\
&= \frac{1}{4} \left [\frac{\Gamma (\frac{1}{4}) \Gamma (\frac{1}{2})}{\Gamma (\frac{3}{4})} - \frac{\Gamma (\frac{3}{4}) \Gamma (\frac{1}{2})}{\Gamma (\frac{1}{4})} \right ] \tag3\\
&= \frac{\Gamma^2 (\frac{1}{4})}{4 \sqrt{2 \pi}} - \frac{\pi \sqrt{2 \pi}}{\Gamma^2 (\frac{1}{4})}.\tag4
\end{align}
Explanation
(2) Each integral is of the form of the Beta function given in (1).
(3) Using the result $\operatorname{B}(x,y) = \dfrac{\Gamma (x) \Gamma (y)}{\Gamma (x + y)}$. 
(3) Using $\Gamma (\frac{1}{2}) = \sqrt{\pi}$ and $\Gamma (\frac{3}{4}) = \dfrac{\pi \sqrt{2}}{\Gamma (\frac{1}{4})}$, a result that comes immediately from Euler's reflexion formula for the Gamma function.

If it is the indefinite integral you are really after, then as suggested in the comments, your integral can be expressed in terms of elliptic functions of the first and second kinds.
From ($*$), namely
$$I = \frac{1}{2} \int \frac{du}{\sqrt{1 - u^2} \sqrt{u}} - \frac{1}{2} \int \frac{\sqrt{u}}{\sqrt{1 - u^2}} \, du,$$
where $u = x^2$, substituting $u = \sin \theta$ into each of the above integrals one arrives at
\begin{align}
I &= \frac{1}{2} \int \sqrt{\operatorname{cosec} x} \, d\theta - \frac{1}{2} \int \sqrt{\sin \theta} \, d\theta\\
&= - F \left (\frac{\pi - 2 \theta}{4} \Big{|} 2 \right ) + E \left (\frac{\pi - 2 \theta}{4} \Big{|} 2 \right ) + C\\
&= E \left (\frac{\pi - 2 \sin^{-1} (x^2)}{4} \Big{|} 2 \right ) - F \left (\frac{\pi - 2 \sin^{-1} (x^2)}{4} \Big{|} 2 \right ) + C, \qquad -1 \leqslant x \leqslant 1.
\end{align}
Here $F( \phi | k^2)$ is an elliptic integral of the first kind while $E(\phi | k^2)$ is an elliptic integral of the second kind.
