How can I find $\int_{\sqrt2/2}^{1}\int_{\sqrt{1-x^2}}^{x}\frac{1}{\sqrt{x^2+y^2}}dydx$? My question is ; How can I solve the following integral question?
$\displaystyle \int_{\sqrt2/2}^{1}\int_{\sqrt{1-x^2}}^{x}\frac{1}{\sqrt{x^2+y^2}}dydx$
Thanks in advance,

You've had some time to study this, so let's look closer at the two evident approaches:  (a) conversion to polar coordinates, (b) integrate directly with a standard hyperbolic substitution.
(a) Conversion to polar coordinates: Let
$$I = \int_{1/\sqrt{2}}^1 \, \int_{\sqrt{1-x^2}}^x \frac{1}{\sqrt{x^2+y^2}} \text{d}y \, \text{d}x.$$
If you draw a diagram of the situation you will see that
$$I = \int_0^{\pi/4} \, \int_{r=1}^{r=\sec \theta} \frac{1}{r} r \text{d}r \, \text{d} \theta = \int_0^{\pi/4} \sec \theta \, - 1 \textrm{ d} \theta $$
$$= \left[ \frac{1}{2} \log \left| \frac{1+ \sin \theta}{1- \sin \theta} \right| - \theta  \right]_0^{\pi/4} = \log(1+\sqrt{2}) - \frac{\pi}{4}.$$
(b) Standard hyperbolic substitution: To evaluate the inner integral set $y= x \sinh u$ and noting that $\sinh^{-1} u = \log(u+ \sqrt{1+u^2})$ we have
$$\int_{\sqrt{1-x^2}}^x \frac{1}{\sqrt{x^2+y^2}} \text{d}y  =
\left[ \log \left( \frac{y}{x} + 
\sqrt{ 1+ \frac{y^2}{x^2} }\right) \right]_{\sqrt{1-x^2}}^x $$
$$= \log(1+ \sqrt{2}) + \log x - \log( 1+ \sqrt{1-x^2}). \quad (1)$$
Both the logs can be integrated by parts. The first is standard, $\int \log x \textrm{ d}x = x \log x - x + C$ and the second
$$ \int \log( 1+ \sqrt{1-x^2} ) \textrm{ d}x = x \log( 1+ \sqrt{1-x^2}) -x + \sin^{-1} x + C.$$
You will need (do the integration) to note that
$$\frac{1}{\sqrt{1-x^2}} - 1 = \frac{x^2}{(1-x^2) + \sqrt{1-x^2}}.$$
And so upon integrating $(1)$ between $1/\sqrt{2}$ and $1$ you obtain
$$I = \left(1 - \frac{1}{\sqrt{2}} \right) \log(1+ \sqrt{2}) +
\left[ x \log x - x\log( 1+ \sqrt{1-x^2}) - \sin^{-1} x \right]_{1/\sqrt{2}}^1$$
$$= \log(1+\sqrt{2}) - \frac{\pi}{4}.$$
