Integrating $\int{\frac{\sqrt{1-x^2}}{(x+\sqrt{1-x^2})^2} dx}$ I am a little bit lost with integral: $$\int{\frac{\sqrt{1-x^2}}{(x+\sqrt{1-x^2})^2} dx}$$
I have already worked on in and done substitution $x = \sin(t)$:
This brings me to: $$\int{\frac{\cos(t)^2}{(\sin(t)+\cos(t))^2}dt}$$
Further treating denominator to achieve:
$$\int{\frac{\cos(t)^2}{\sin(2t)+1}dt}$$
I can split this fraction into two integrals by doing $\cos(t)^2 = 1-\sin(t)^2$ but this doesn't help me to solve the integral further. 
Please, can you show me how to continue to "break it" :-)
Best thanks!

Instead of doing a trig substitution, expand the denominator and simplify the integrand.
