Question

Show that $\int \frac{x\sqrt{a^2-x^2}}{a^2+x^2}dx=\sqrt{a^2-x^2}-\frac{a}{\sqrt{\left(2\right)}}\log \frac{a\sqrt{2}+\sqrt{a^2-x^2}}{a\sqrt{\left(2\right)}-\sqrt{(a^2-x^2)}}$.

Answer 1

$I=\int \frac{x\sqrt{a^2-x^2}}{a^2+x^2}dx$
$x=a\sin t\Rightarrow dx=a\cos tdt$
$I=a\int \frac{a^2\sin t{\cos }^{2}t}{a^2{\sin }^{2}t+a^2}dt=a^3\int \frac{\sin t{\cos }^{2}t}{a^2{\sin }^{2}t+a^2}dt$
$=a^3\int -\frac{\sin t{\cos }^{2}t}{a^2({\cos }^{2}t-2)}dt$
Now put $u=\cos t\Rightarrow du-\sin tdt$
$I=a\int \frac{u^2}{u^2-2}du=a\int (\frac{2}{u^2-2}+1)du$
$=-a\int \frac{1}{1-\frac{u^2}{2}}du+a\int du$
$=-\frac{a}{\sqrt{2}}\log \left(\frac{a\sqrt{2}+\sqrt{a^2-x^2}}{a\sqrt{2}-\sqrt{a^2-x^2}}\right)+\sqrt{a^2-x^2}$