How to integrate this $\int\frac{\mathrm{d}x}{{(4+x^2)}^{3/2}} $ without trigonometric substitution? I have been looking for a possible solution but they are with trigonometric integration..
I need a solution for this function without trigonometric integration
$$\int\frac{\mathrm{d}x}{{(4+x^2)}^{3/2}}$$
 A: Let $I=\int \frac{dx}{\sqrt{x^2+4}} \,;\, J=\int \frac{dx}{\sqrt{x^2+4}^3}$
We integrate $I$ by parts, we get:
$u=(x^2+4)^{-1/2}, dv=dx$
$du=-x(x^2+4)^{-3/2}, v=x$
Thus
$$I= \frac{x}{\sqrt{x^2+4}}+ \int \frac{x^2}{\sqrt{x^2+4}^3}= \frac{x}{\sqrt{x^2+4}}+ \int \frac{x^2+4-4}{\sqrt{x^2+4}^3}=$$
$$= \frac{x}{\sqrt{x^2+4}}+ I-4J=$$
Thus, canceling the $I$ we get
$$4J=  \frac{x}{\sqrt{x^2+4}} +C$$
A: $$\int\frac{dx}{(a^2+x^2)^{\frac n2}}=\int1\cdot \frac1{(a^2+x^2)^{\frac n2}}dx$$
$$=\frac x{(a^2+x^2)^{\frac n2}}-\int\left(\frac{-n}2\frac{2x\cdot x}{(a^2+x^2)^{\frac n2+1}} \right) dx$$
$$=\frac x{(a^2+x^2)^{\frac n2}}+n\int \left(\frac{(a^2+x^2-a^2)}{(a^2+x^2)^{\frac n2+1}}\right)dx $$
$$=\frac x{(a^2+x^2)^{\frac n2}}+n\left(\int\frac{dx}{(a^2+x^2)^{\frac n2}}-a^2\int\frac{dx}{(a^2+x^2)^{\frac n2+1}}\right)+c $$ where $c$ is the constant for indefinite integration.
or, $$na^2\int\frac{dx}{(a^2+x^2)^{\frac n2+1}}=\frac x{(a^2+x^2)^{\frac n2}}+(n-1)\int\frac{dx}{(a^2+x^2)^{\frac n2}}+c$$
Putting $n=1,$  $$a^2\int\frac{dx}{(a^2+x^2)^{\frac 32}}=\frac x{(a^2+x^2)^{\frac 12}}+c$$
A: $$\frac{1}{\left(4+x^2\right)^{3/2}}=\frac{1}{8}\cdot\frac{1}{\left(1+\left(\frac{x}{2}\right)^2\right)^{3/2}}$$
Now try 
$$x=2\sinh u\implies dx=2 \cosh u\,du\implies$$
$$\int\frac{dx}{\left(4+x^2\right)^{3/2}}=\frac{1}{8}\int\frac{2\,du\cosh u}{(1+\sinh^2u)^{3/2}}=\frac{1}{4}\int\frac{du}{\cosh^2u}=\ldots $$
A: The following is not quite right, it needs minor modification for negative $x$. Let $x=\dfrac{1}{t}$. Then $dx=-\dfrac{1}{t^2}$. Substitute and do some algebra. There is some nice cancellation, and we end up with
$$\int \frac{-t\,dt}{(4t^2+1)^{3/2}}.$$
Now let $u=4t^2+1$. 
