Q:

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

A:

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 
$$\int \frac {dx}{(b^2+a^2 x^2)^\frac 3 2} = \frac 1 {b^2 }  \frac x {\sqrt{b^2+a^2x^2)}}.$$ 

