Hint for solving a indefinite integral

Question

Can anyone provide a hint for solving this definite integral $$\int \dfrac{\sqrt{1-a^2 +x^2}}{x^2(a^2-x^2)}dx$$ Here $a$ is a real constant.

I'll appreciate any help.

Thanks.

Answer 1

Let be $$ I=\int \frac{\sqrt{1-a^2+x^2}}{x^2(a^2-x^2)}\mathrm d x $$ with the substitution $ \frac{\sqrt{1-a^2+x^2}}{x}=u $ and $x^2=\frac{1-a^2}{u^2-1}$ and $\mathrm d u=\frac{a^2-1}{x^2\sqrt{1-a^2+x^2}}\mathrm d x$ and then $\frac{x^2\sqrt{1-a^2+x^2}}{a^2-1}\mathrm d u=\mathrm d x$ the integral $I$ becomes

$$ J=\int \frac{u^2}{1-u^2a^2}\mathrm d u $$ and with the substitution $au=t$ the integral $J$ becomes $$ K=\frac{1}{a^3}\int \frac{t^2}{1-t^2}\mathrm d t=\begin{cases}\frac{1}{a^3}\left[\tanh^{-1}(t)-t\right]+c & \text{for }|t|<1\\ \frac{1}{a^3}\left[\coth^{-1}(t)-t\right]+c & \text{for }|t|>1 \end{cases} $$ observing that $\left(\tanh^{-1}(t)\right)'=\frac{1}{1-t^2}$ for $|t|<1$ and $\left(\coth^{-1}(t)\right)'=\frac{1}{1-t^2}$ for $|t|>1$ Thus $J=K(au)$ and $I=K\left(a\frac{\sqrt{1-a^2+x^2}}{x}\right)$ $$ I=\begin{cases}\frac{1}{a^3}\left[\tanh^{-1}\left(a\frac{\sqrt{1-a^2+x^2}}{x}\right)-a\frac{\sqrt{1-a^2+x^2}}{x}\right]+c & \text{for }\left|a\frac{\sqrt{1-a^2+x^2}}{x}\right|<1\\ \frac{1}{a^3}\left[\coth^{-1}\left(a\frac{\sqrt{1-a^2+x^2}}{x}\right)-a\frac{\sqrt{1-a^2+x^2}}{x}\right]+c & \text{for }\left|a\frac{\sqrt{1-a^2+x^2}}{x}\right|>1 \end{cases} $$ or $$ I=\begin{cases}\frac{1}{a^3}\left[\tanh^{-1}\left(a\frac{\sqrt{1-a^2+x^2}}{x}\right)-a\frac{\sqrt{1-a^2+x^2}}{x}\right]+c & \text{for }\left|a\frac{\sqrt{1-a^2+x^2}}{x}\right|<1\\ \frac{1}{a^3}\left[\tanh^{-1}\left(\frac{x}{a\sqrt{1-a^2+x^2}}\right)-a\frac{\sqrt{1-a^2+x^2}}{x}\right]+c & \text{for }\left|a\frac{\sqrt{1-a^2+x^2}}{x}\right|>1 \end{cases} $$ observing that $\tanh^{-1}\left(\frac{1}{z}\right)=\coth^{-1}(z)$.