How to integrate $1/\sqrt{(1+x^2)^3}$? Normally I use WolframAlpha pro to help me with problems I don't know however wolfram wont/cant show me the steps only the final solution to this integration problem. 
Is anyone able to assist me with a walk through of atleast the start if not all of the steps to solving this equation?
$$\int\frac{dx}{\sqrt{(1+x^2)^3}}$$

$$\int\frac{dx}{\sqrt{(1+x^2)^3}}$$
make a trigonometric substituition
$$
x=\tan\theta\\
dx=\sec^2\theta d\theta\\
1+x^2=\sec^2\theta\\
\int\frac{\sec^2\theta}{\sqrt{(\sec^2\theta)^3}}d\theta$$
if $\theta\in(0|\frac{\pi}{2})\cup(\frac{3\pi}{2}|2\pi)$ then $\cos\theta>0$ and then
$$\begin{align}
\tan\theta&=x\\
\sec\theta&=\sqrt{1+x^2}\\
\cos\theta&=\frac{1}{\sqrt{1+x^2}}\\
\sin\theta&=\frac{x}{\sqrt{1+x^2}}\\
\int\frac{\sec^2\theta}{\sqrt{(\sec^2\theta)^3}}d\theta&=\int\frac{\sec^2\theta}{\sec^3\theta}d\theta\\
&=\int\frac{1}{\sec\theta}d\theta\\
&=\int\cos\theta d\theta\\
&=\sin\theta+C\\
&=\frac{x}{\sqrt{1+x^2}}+C
\end{align}$$
if $\theta\in(\frac{\pi}{2}|\frac{3\pi}{2})$ then $\cos\theta<0$ and then
$$\begin{align}
\tan\theta&=x\\
\sec\theta&=-\sqrt{1+x^2}\\
\cos\theta&=-\frac{1}{\sqrt{1+x^2}}\\
\sin\theta&=-\frac{x}{\sqrt{1+x^2}}\\
\int\frac{\sec^2\theta}{\sqrt{(\sec^2\theta)^3}}d\theta&=\int-\frac{\sec^2\theta}{\sec^3\theta}d\theta\\
&=\int-\frac{1}{\sec\theta}d\theta\\
&=\int-\cos\theta d\theta\\
&=-\sin\theta+C\\
&=\frac{x}{\sqrt{1+x^2}}+C
\end{align}$$
