$\int\frac{1}{x^2\sqrt{a^2-x^2}}\mathrm{d}x$ via Partial Fractions

Question

I was wondering how to compute integrals of the form \begin{align} \int\frac{1}{x^2\sqrt{a^2-x^2}}\mathrm{d}x \end{align} using partial fractions. This question comes from Strang and Herman's Calculus, Volume 2. I know the standard method would be trigonometric substitution, but the question explicitly states to use a substitution to convert the integrand to a rational function. (I also feel obliged to mention this is not a homework problem.)

Answer 1

For the laughs, note that under $t=\frac{\sqrt{a^2-x^2}}{x}$ , $$\int\frac{dx}{x^2\sqrt{a^2-x^2}}=\int\frac{1}{\frac{a^3t}{(t^2+1)^{3/2}}}\left(-\frac{at}{(t^2+1)^{3/2}}dt\right)\\=-\frac{1}{a^2}\int \color{red}1 dt=-\frac{1}{a^2}\int\color{red}{\frac{t^2}{t^2}}dt=-\frac{1}{a^2}\int\color{red}{1+\frac0t}dt$$

Answer 2

Okay, the substitution seems to be $x=\frac{2at}{1+t^2}$ as suggested by guoran guan. If one applies the standard trigonometric substitution $x=a\sin\theta$, transforming the integral as \begin{align} \int\frac{1}{x^2\sqrt{a^2-x^2}}\mathrm{d}x=\frac{1}{a^2}\int\frac{1}{\sin(\theta)^2}\mathrm{d}\theta, \end{align} then one can apply the Weierstrass substitution as hinted by coffeemath; explicitly, we have \begin{align} \frac{1}{a^2}\int\frac{1}{\sin(\theta)^2}\mathrm{d}\theta&=\frac{1}{a^2}\int\frac{1}{\bigl(\frac{2t}{1+t^2}\bigr)^2}\frac{2}{1+t^2}\mathrm{d}t\\ &=\frac{1}{a^2}\int\frac{1+t^2}{2t^2}\mathrm{d}t \end{align} which then can be done by partial fractions.