Evaluate $\int \frac{dx}{(b^2 + a^2x^2)^{3/2}}$ where $a,b \in \mathbb R^+$ 
Evaluate $$\int \frac{dx}{(b^2 + a^2x^2)^{3/2}}$$ where $a,b \in \mathbb R^+$

I let $(ax) = b\tan(\theta)$, then $dx = \frac{b}{a}\sec^2(\theta)$
$$= \int \frac{b\sec^2(\theta)}{ab^3(\sec^2(\theta))^{3/2}}$$
$$ = \frac{1}{ab^2} \int \frac{1}{\sec(\theta)}d\theta $$
$$= \frac{1}{ab^2}\sin(\theta) + C$$
$$= \frac{1}{ab^2} \frac{ax}{\sqrt{(ax)^2+b^2}} + C$$
Is this right? Other website did it differently not sure if its right


Yes, it is true!

You can always verify your integrals by differentiating them.

The integral can also be solved without using trigonometric functions.
Instead, substitute $ax=by$.
Then  $$\int \frac {dx}{(b^2+a^2 x^2)^\frac 3 2} = \frac 1 {ab^2 } \int \frac {dy} {(1+y^2)^\frac 3 2}.$$ 
This integral can be solved as follows:
Observe that 
$$
\frac 1 {(1+y^2)^\frac 3 2}=\frac {1+y^2}{(1+y^2)^\frac 3 2}-\frac {y^2} {(1+y^2)^\frac 3 2}
= \frac {1}{(1+y^2)^\frac 1 2}-\frac {y^2} {(1+y^2)^\frac 3 2}. 
$$
Since $$\frac d {dy} \frac 1 {\sqrt{1+y^2}}= -\frac y {(1+y^2)^\frac 3 2}
$$ 
we can use integration by parts to get 
$$
\int \frac {y^2} {(1+y^2)^\frac 3 2} \,dy 
=-\int y \,\frac d {dy}\frac 1 {\sqrt{1+y^2}}\, dy
=-\frac y {\sqrt{1+y^2}} + \int \frac 1 {\sqrt{1+y^2}} \, dx.
$$ 
This implies 
$$
\int \frac 1 {(1+y^2)^\frac 3 2} = \frac y {\sqrt{1+y^2}}
$$
Subsituting back and multiplying with the factor $\frac 1 {ab^2}$, we get the desired result 
$$\int \frac {dx}{(b^2+a^2 x^2)^\frac 3 2} = \frac 1 {b^2 }  \frac x {\sqrt{b^2+a^2x^2)}}.$$ 
