Evaluation of $\int_{0}^{1} \frac{1}{x+\sqrt{1-x^2}} \space dx$ Evaluate
$$\int_{0}^{1} \frac{1}{x+\sqrt{1-x^2}} \space dx$$

My main concern is finding the indefinite integral as once i have that the rest is fairly straight forward. Please give a detailed answer with reference to why you made each substitution (what indicated that said substitution would work etc.)

My initial substitution was $x= \sin \theta$ which tidies it up a bit

$$\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sin x + \cos x} \space dx$$

Also the answer from Wolfram Alpha is;

$$ = \frac{1}{4}\bigg( log(1-2 x^2)+2 \tanh^{-1}\bigg(\frac{x}{\sqrt{1-x^2}}\bigg)+2 \sin^{-1}x\bigg)+constant$$

I need to be able to answer questions like this in under 15 mins so any help is appreciated
There's a simple solution somewhere in M.SE which goes like this...

$$I = \int_0^{\pi/2} dt \frac{\cos{t}}{\cos{t}+\sin{t}} $$
$$J = \int_0^{\pi/2} dt \frac{\sin{t}}{\cos{t}+\sin{t}} $$

$$I + J = \frac{\pi}{2}$$
$$I - J = \int_0^{\pi/2} dt \frac{\cos{t}-\sin{t}}{\cos{t}+\sin{t}} = \int_0^{\pi/2} dt \frac{(\cos{t}+\sin{t})'}{\cos{t}+\sin{t}}=[\log{(\cos{t}+\sin{t})}]_0^{\pi/2} = 0$$

Thus $I = J = \frac{\pi}{4}$.

ADDENDUM

The above technique requires the easy computation of the antiderivative, which you should see is

$$\frac12 \left [t + \log{(\cos{t}+\sin{t})} \right ] + C = \frac12 \left [\arcsin{x} + \log{(x+\sqrt{1-x^2})} \right ] + C$$
Finding indefinite integral:

\begin{align}
&amp;x\mapsto\sin u\\
&amp;I=\int \frac{\cos u}{\sin u + \cos u}du\\
&amp;=\frac12\int \frac{\cos u - \sin u + \cos u + \sin u}{\sin u + \cos u}du=\frac12 \int\frac{\cos u-\sin u}{\sin u + \cos u}du+\frac u2\\
&amp;=\frac12 \ln (\cos u + \sin u) + \frac u2+C\\
&amp;=\frac12 \ln (\sqrt{1-x^2} + x) + \frac {\sin^{-1}x}{2}+C\\
\end{align}