Can someone help me with this integration question $$\int_0^1 \frac{dx} {x+\sqrt{1-x^2}} $$
I tried substitution with x=sinx and also tries times both denominator and numerator by $$ x- \sqrt {1-x^2} $$ but it becomes even more complicated. Can someone gives me a hint how to solve this question please?
This is just another variation on a theme that has been used in other answers, but I think it might be worth seeing in this guise:
$$
\begin{align}
\int_0^1\frac{\mathrm{d}x}{x+\sqrt{1-x^2}}
&=\int_0^1\frac{x}{\sqrt{1-x^2}}\frac{\mathrm{d}x}{x+\sqrt{1-x^2}}\tag{1}\\
&=\frac12\int_0^1\frac{\mathrm{d}x}{\sqrt{1-x^2}}\tag{2}\\
&=\frac12\int_0^{\pi/2}\mathrm{d}\theta\tag{3}\\[3pt]
&=\frac\pi4\tag{4}
\end{align}
$$
Explanation:
$(1)$: substitute $x\mapsto\sqrt{1-x^2}$
$(2)$: average the left and right sides of $(1)$
$(3)$: substitute $x\mapsto\sin(\theta)$
$(4)$: integrate
Set $x=\sin y$
Then apply $$I=\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$$
$$I+I=\int_a^bf(x)\ dx+\int_a^bf(a+b-x)\ dx$$
Your second approach gets the denominator to be $x^2 - (1- x^2) = 2x^2 - 1$, and you can split the numerator into two parts, so you have to separate integrals. That's good.
Now do you deal with that denominator? Try $x = u / \sqrt{2}$. Then you can do an sine or cosine substitution (or even secant) substitution.
First Method
$1)$ We make subtitution $x = \sin (u)$ and conclude that
$$I = \int _0^{{\pi \over 2}}{{\cos \left( u \right)} \over {\sin \left( u \right) + \cos \left( u \right)}}\;du\tag{1}$$
$2)$ Next, we use this substitution $u=\pi / 2 - v$ to get
$$I=\int _0^{{\pi \over 2}}{{\sin \left( v \right)} \over {\cos \left( v \right) + \sin \left( v \right)}}\;dv\tag{2}$$
$3)$ Sum the previous results
$$2I = \int_0^{{\pi \over 2}} {{{\sin (u) + \cos \left( u \right)} \over {\sin \left( u \right) + \cos \left( u \right)}}du} = \int_0^{{\pi \over 2}} {1du} = {\pi \over 2}\,\,\,\,\, \to \,\,\,\,I = {\pi \over 4}\tag{3}$$
Second Method
$1)$ Use the tangent half angle substitution, $u={\tan(\frac{v}{2})}$, in equation $(1)$ to get
$$I=\int _0^\infty {1 \over {\left( {{v^2} + 1} \right)\left( {v + 1} \right)}}\;dv\tag{4}$$
$2)$ Use partial fraction to decompose the integrand and finally
$$\eqalign{ & I = {1 \over 2}\int_0^\infty {\left( {{1 \over {v + 1}} + {{ - v} \over {{v^2} + 1}} + {1 \over {{v^2} + 1}}} \right)dv} \cr & \,\,\, = \left. {{1 \over 2}\left( {\ln (v + 1) - {1 \over 2}\ln ({v^2} + 1) + \arctan (v)} \right)} \right|_0^\infty \cr & \,\,\, = \mathop {\lim }\limits_{v \to \infty } {1 \over 2}\left( {\ln (v + 1) - {1 \over 2}\ln ({v^2} + 1) + \arctan (v)} \right) = {\pi \over 4} \cr} \tag{5}$$