Q: $\displaystyle \int \frac{1}{x\sqrt{1-x^2}} \ \mathrm{d}x$
Is there something really simple that I am not seeing? How would this be done? Thanks in advance.
Hello,
You can make a substitution $\displaystyle t=\sqrt{1-x^2}$
$\displaystyle \implies dt=\frac{-x}{t} \ dx \implies dx=-\frac tx \ dt$
$\displaystyle \int \frac{1}{x \sqrt{1-x^2}} dx=\int \frac{-1}{x^2} dt=- \int \frac{1}{1-t^2} dt=-\frac 12 \int \frac{1}{1-t}+\frac{1}{1+t} dt$
(modulo my mistakes )