Have you editted the first post? I didn't notice that "x" outside the square root before.
Let u= x2- b2 so that du= 2xdx
\sqrt{\frac{b^2-a^2}{x^2- b^2}-1}= \sqrt{\frac{A}{u}- 1}
#7
WiFO215
416
1
Yes. I edited the first post. Forget it though. I think what I did is wrong anyway.