How to integrate $\int\frac{dx}{(A^2+x^2)^{3/2}}$ I have the following to integrate:
$$\int\frac{dx}{(A^2+x^2)^{3/2}}$$
Where, $A$ is a constant.
Not really sure what technique to use. Any hint would be helpful.
 A: let $x=A\tan\theta\implies dx=A\sec^2\theta\ d\theta$
$$\int \frac{1}{(A^2+x^2)^{3/2}}dx$$
$$=\int \frac{1}{(A^2+A^2\tan^2\theta)^{3/2}}(A\sec^2\theta\ d\theta)$$
$$=\int \frac{A\sec^2\theta\ d\theta}{A^{3}\sec^3\theta}$$
$$=\frac{1}{A^2}\int \cos \theta\ d\theta$$
$$=\frac{1}{A^2}\sin\theta+C$$
$$=\frac{1}{A^2}\sin\left(\tan^{-1}\frac{x}{A}\right)+C$$
$$=\frac{1}{A^2}\sin\left(\sin^{-1}\frac{x/A}{\sqrt{1+(x/A)^2}}\right)+C$$
$$=\color{red}{\frac{x}{A^2\sqrt{A^2+x^2}}+C}$$
A: Hint: Use the trigonometric substitution $x=A\tan u$ and $dx=A \sec^2 u$.
A: Use substitution : $x=\tan t$. You get the integral 
$$\displaystyle\int\frac{\mathrm d\mkern1.5mu t}{A^2(1+\tan^2t)^2}=\int \cos t\,\mathrm d\mkern1.5mu t=\frac1{A^2}\sin t=\frac1{A^2}\sin(\arctan t)=\frac{x}{A^2\sqrt{1+x^2}}.$$
A: 
$$\int\frac{dx}{(A^2+x^2)^{3/2}}$$

Hint:
Substitute $x=A\tan(t)$ and $dx=A\sec^2(t)dt$. Then $(x^2+A^2)^{3/2}=(A^2\tan^2(t)+A^2)^{3/2}=A^3\sec^3(t)$ and $t=\arctan(\frac x A)$
Full answer:

$$A\int\frac{\cos t}{A^3}dt=\frac{1}{A^2}\int \cos t dt=\frac{\sin t}{A^2}+\mathcal C=\frac{1}{A^2}\sin\left(\arctan(x/A)\right)+\mathcal C=\color{red}{\frac{x}{A^2\sqrt{x^2+A^2}}+\mathcal C}$$

A: The substitution $$u=\frac{1}{(a^2+x^2)^\frac{1}{2}}$$ transforms the integral to $$-\frac{1}{a^2}\int \frac{au}{\sqrt{1-a^2u^2}}d(au)=\frac{\sqrt{1-(au)^2}}{a^2}$$ Upon resubstitution the last term is seen to be same as $$\frac{x}{a^2\sqrt{a^2+x^2}}$$
A: This is actually a simple integral. Use the substitution
$$x = A\tan (y) ~~~~~~~ \text{d}x = A\sec^2(y)\ \text{d}y$$
In this way
$$(A^2 + x^2)^{3/2} = (A^2\tan^2(y) + A^2)^{3/2} = A^3\sec^3(y)$$
and $y = \arctan(x/A)$
So your integration becomes
$$\frac{1}{A^2}\int\cos(y)\ \text{d}y$$
which is trivial.
Final answer:
$$\frac{\sin(\arctan(x/A))}{A^2}$$
