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 $$\begin{align} \tan\theta&=x\\ \sec\theta&=-\sqrt{1+x^2}\\ \cos\theta&=-\frac{1}{\sqrt{1+x^2}}\\ \sin\theta&=-\frac{x}{\sqrt{1+x^2}}\\ \int\frac{\sec^2\theta}{\sqrt{(\sec^2\theta)^3}}d\theta&=\int-\frac{\sec^2\theta}{\sec^3\theta}d\theta\\ &=\int-\frac{1}{\sec\theta}d\theta\\ &=\int-\cos\theta d\theta\\ &=-\sin\theta+C\\ &=\frac{x}{\sqrt{1+x^2}}+C \end{align}$$