My question is that I want to evaluate step-by-step the integral
$$\int_0^1 \frac{\mathrm{d}x}{(a^2+x^2)\sqrt{1-x^2}}$$
where $a$ is some constante in the real numbers and $x$ is also a real number.
My attempt: If $x = \sin(u)$ then the integral becomes
$$\frac{1}{a^2}\int_0^{\arcsin(1)}\frac{\mathrm{d}u}{1+(\sin(u)/a)^2}$$
Then I can get to other places but this isn't quite good. Like using another change of the variables. From wolfram this integral get's in to a $\tan^{-1}$ function.
With $x=\sin u$ we have,
$$\int_{0}^{\frac{\pi}{2}} \frac{1}{\sin^2(u)+a^2} du$$
Dividing top and bottom by $\cos^2 u$ gives,
$$\int_{0}^{\frac{\pi}{2}} \frac{\sec^2 u}{\tan^2 u+a^2 \sec^2 u} du$$
$$=\int_{0}^{\frac{\pi}{2}} \frac{\sec^2 u}{\tan^2 u+a^2(\tan^2 u+1)} du$$
$$=\int_{0}^{\frac{\pi}{2}} \frac{\sec^2 u}{(a^2+1)\tan^2 u+a^2} du$$
Now with $\tan u=v$ we have,
$$\int_{0}^{\infty} \frac{1}{(a^2+1)v^2+a^2} dv$$
I think you can handle this.
Hint...when you do the substitution $x=\sin u$ you get $$I=\int_0^{\frac{\pi}{2}}\frac{1}{a^2+\sin^2u}du$$
Now do the substitution $t=\tan u$ and the integral becomes$$I=\int_0^{\infty}\frac{1}{a^2+(a^2+1)t^2}dt$$
Can you finish this?