<s>[INST] Find the value of this indefinite integral. $$
\int \frac{dx}{(a^2+x^2)\sqrt{l^2+a^2+x^2}}
$$
where both $a$ and $l$ are constants. 
I've tried simplifying :
$$
\int \frac{dx}{(a^2+x^2)\sqrt{l^2+a^2+x^2}}
\\
=\frac{1}{l^2}\int \frac{(l^2+a^2+x^2-a^2-x^2)dx}{(a^2+x^2)\sqrt{l^2+a^2+x^2}}
\\
=\frac{1}{l^2}\int \frac{(\sqrt{l^2+a^2+x^2})dx}{(a^2+x^2)}-\frac{1}{l^2}\int \frac{dx}{\sqrt{a^2+x^2+l^2}}
$$
Now, I know how to simplify the second integral but I have no idea how to solve the first one. 
Note : I need the value of this integral for a physics problem. 
 [/INST] Take $l^2+a^2 = r^2$ and $x=r\tan (\theta)$ so that the indefinite integral becomes $$\int \frac{r \sec^2(\theta) d\theta}{(r \sec(\theta))(r^2 \tan^2(\theta) + a^2)} = \int \frac{\sec(\theta) d\theta}{r^2 \sec^2(\theta) - l^2} = \int \frac{\cos(\theta) d\theta}{r^2-l^2 \cos^2(\theta)} =$$ $$\int \frac{\cos (\theta) d\theta}{(r^2-l^2)+l^2\sin^2(\theta)} = \int \frac{\cos(\theta) d\theta}{a^2+l^2 \sin^2(\theta)}$$Now take $\sin(\theta)=y$ so that the integral becomes 
$$\int \frac{dy}{a^2+l^2y^2}$$Now take $y=\frac{a}{l}t$ and the integral becomes $$\frac al \int \frac{dt}{a^2(1+t^2)}=\frac 1{al}\int \frac{dt}{1+t^2} = \frac{1}{al} \tan^{-1}(t)=\frac{1}{al} \tan^{-1} \left( \frac la \left( \sin(\theta) \right) \right)$$ $$=\frac{1}{al} \tan^{-1} \left( \frac la \left( \frac{x}{\sqrt{x^2+r^2}} \right) \right)=\frac{1}{al} \tan^{-1} \left( \frac la \left( \frac{x}{\sqrt{x^2+a^2+l^2}} \right) \right)$$ as @innerproduct mentioned in his answer.
</s>