The value of the integral $\int\limits_0^a \frac{1}{x+\sqrt{a^2-x^2}} d x$, is

$\frac{\pi}{4}$

Let $I=\int\limits_0^a \frac{1}{x+\sqrt{a^2-x^2}} d x$

Putting $x=a \sin \theta$, we get

$I=\int\limits_0^{\pi / 2} \frac{\cos \theta}{\sin \theta+\cos \theta} d \theta$         .....(i)

$\Rightarrow I =\int\limits_0^{\pi / 2} \frac{\sin \theta}{\cos \theta+\sin \theta} d \theta $        [Using $\int\limits_0^a f(x)dx = \int\limits_0^a f(a-x)dx$]

Adding (i) and (ii), we get

$2 I=\int\limits_0^{\pi / 2} 1 d \theta=\frac{\pi}{2} \Rightarrow I=\frac{\pi}{4}$