Integral $\int \frac{ax^2-b}{x\sqrt{c^2x^2-(ax^2+b)^2}} dx$ Integrate:
$$\int \frac{ax^2-b}{x\sqrt{c^2x^2-(ax^2+b)^2}} dx$$
What should be the substitution here so that it becomes simpler??? Please help
 A: This question follows a series of substitution, I'd say it's very tedious. However, yet simple.
Here's a hint: 
First substitute. $x^2=t$, giving $dx=\dfrac{dt}{2x}$
Giving us, the integral as :  
$$\large\int\dfrac{(at-b)}{2t\sqrt{c^2t-(at+b)^2}}.dt$$  
The same can be re written as,  
$$\large\int\dfrac{a}{2\sqrt{c^2t-(at+b)^2}}.dt-\large\int\dfrac{b}{2t\sqrt{c^2t-(at+b)^2}}.dt$$  
Considering the first integral. 
Put $at+b=y$ , so that $adt=dy$ and $t=\dfrac{y-b}{a}$.
After all the substitution, you should get.  
$I_1=\dfrac{1}{2}\large\int\dfrac{1}{\sqrt{c^2\left(\dfrac{y-b}{a}\right)-(y)^2}}.dy$  
Which is a rational function, and can be easily solved by completing the square.(try!)
End result should be(hoping no careless mistake!)  

$~~~~~~~~~~~~~~~~~$$\begin{equation}I_1=\dfrac{1}{2}\sin^{-1}\left(\dfrac{(ax^2+b)-\dfrac{c^2}{2a}}{\left(\left(\dfrac{c^2}{2a}\right)^2-\dfrac{c^2b}{a}\right)^{\dfrac{1}{2}}}\right) \end{equation}$  $~~~~~~~~~~~~~~~~~~~$ Iff, $\left(\dfrac{c^2}{2a}\right)^2>\dfrac{c^2b}{a}$   

For integral $I_2$ or the second integral. Procedure is fairly similar.  
The only difference is that first you have to put $t=\dfrac{\large{1}}{\large{z}}$. Then the procedure is same as first integral giving you.  

$~~~~~~~~~~~~~~~~~$$\begin{equation}I_2=\dfrac{1}{2}\sin^{-1}\left(\dfrac{(a+\dfrac{b}{x^2})-\dfrac{c^2}{2b}}{\left(\left(\dfrac{c^2}{2b}\right)^2-\dfrac{c^2a}{b}\right)^{\dfrac{1}{2}}}\right) \end{equation}$  $~~~~~~~~~~~~~~~~~~~$ Iff, $\left(\dfrac{c^2}{2b}\right)^2>\dfrac{c^2a}{b}$    

