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

Since no one used it, let me consider the case of the antiderivative $$I=\int\frac{\cos x}{\sin x + \cos x} \space dx$$ Now, use the tangent half-angle substitution (Weierstrass substitution) $t=\tan(\frac x 2)$.
We so obtain $$I=\int \frac{2 \left(t^2-1\right)}{t^4-2 t^3-2 t-1}\space dt$$ But $$t^4-2 t^3-2 t-1=(t^2+1)(t-r_1)(t-r_2)\qquad r_{1,2}=1\pm \sqrt 2$$ Using partial fraction decomposition, we have $$\frac{2 \left(t^2-1\right)}{t^4-2 t^3-2 t-1}=\frac{1-t}{t^2+1}+\frac{1}{2 t-2 \sqrt{2}-2}+\frac{1}{2 t+2 \sqrt{2}-2}$$ that is to say $$\frac{2 \left(t^2-1\right)}{t^4-2 t^3-2 t-1}=\frac{1}{1+t^2}-\frac{2t}{1+t^2}+\frac{1}{2 t-2 \sqrt{2}-2}+\frac{1}{2 t+2 \sqrt{2}-2}$$ and each term is easy to integrate. After simplifications, this leads to $$I=\tan ^{-1}(t)+\frac{1}{2} \log \left(\frac{-t^2+2 t+1}{t^2+1}\right)$$ Now, if integration is from $t=0$ to $t=1$, the logarithms disappear and youare just let with $\tan ^{-1}(1)=\frac \pi 4$.
