Integrating a function with substitution Totally forgot how to integrate.
$$ \int \frac{1}{x^2 \sqrt{x^2+4}}dx$$
Just need a tip, for this what would I use to substitute? 
 A: HINT:
Using Trigonometric Substitution  $$x=2\tan\theta$$
A: Hint:
$$
x^2\sqrt{x^2+4}=x^3\sqrt{1+\frac{4}{x^2}}\,,\quad d\left(\frac{1}{x^2}\right)=-\frac{2}{x^3}d x
$$
A: Put $x = 2 \tan t $, then $dx = 2 \sec^2 t dt $. and $\sqrt{x^2 +4} = \sqrt{ 4 \tan^2 t + 4 } = 2 \sec t$ hence,
$$ \int \frac{dx}{x^2 \sqrt{x^2+4}} = \int \frac{2 \sec^2 t dt}{4 \tan^2 t 2 \sec t} = \frac{1}{4} \int \frac{ \sec t dt }{\tan^2 t} = \frac{1}{4} \int \frac{\frac{1}{\cos t}}{\frac{\sin^2t}{\cos^2 t}} = \frac{1}{4} \int \frac{\cos t dt}{\sin^2 t} = \frac{1}{4} \int \frac{d ( \sin t)}{\sin^2 t} = \frac{-1}{4}\frac{1}{\sin t} + C$$
A: $$\int \frac{1}{x^2 \sqrt{x^2+4}}dx = \int \frac{1}{8(\frac{x}{2})^2 \sqrt{(\frac{x}{2})^2+1}}dx= \int \frac{1}{8tan^2(tan^{-1}(\frac{x}{2})) \sqrt{tan^2(tan^{-1}(\frac{x}{2}))+1}}dx$$
$$=\int \frac{1}{8tan^2(tan^{-1}(\frac{x}{2})) \sqrt{tan^2(tan^{-1}(\frac{x}{2}))+1}}dx\frac{\frac{d(tan^{-1}(\frac{x}{2}))}{dx}}{\frac{d(tan^{-1}(\frac{x}{2}))}{dx}}$$
$$=\int \frac{1}{8tan^2(tan^{-1}(\frac{x}{2})) \sqrt{tan^2(tan^{-1}(\frac{x}{2}))+1}}\frac{d(tan^{-1}(\frac{x}{2}))}{\frac{1}{2}\frac{1}{(\frac{x}{2})^2+1}}$$
$$=\int \frac{(\frac{x}{2})^2+1}{4tan^2(tan^{-1}(\frac{x}{2})) \sqrt{tan^2(tan^{-1}(\frac{x}{2}))+1}}d(tan^{-1}(\frac{x}{2})$$
$$=\int \frac{tan^2(tan^{-1}(\frac{x}{2}))+1}{4tan^2(tan^{-1}(\frac{x}{2})) \sqrt{tan^2(tan^{-1}(\frac{x}{2}))+1}}d(tan^{-1}(\frac{x}{2})$$
$$=\int \frac{sec^2(tan^{-1}(\frac{x}{2}))}{4tan^2(tan^{-1}(\frac{x}{2})) sec(tan^{-1}(\frac{x}{2}))}d(tan^{-1}(\frac{x}{2})$$
$$=\int {\frac{1}{4}cot(tan^{-1}(\frac{x}{2}))csc(tan^{-1}(\frac{x}{2}))}d(tan^{-1}(\frac{x}{2})$$
$$=-\frac{1}{4}csc(tan^{-1}(\frac{x}{2}))+C$$
skipping ahead
$$=-\frac{1}{4}\sqrt{1+\frac{1}{tan^2(tan^{-1}(\frac{x}{2}))}}+C$$
$$=-\frac{1}{4}\sqrt{1+\frac{1}{x^2/4}}+C$$
$$=-\frac{\sqrt{x^2+4}}{4x}+C$$
Does everyone see now why we  use substitutions?
