Calculate $\int\frac{1}{(x+1)\sqrt{1+x^2}} dx $ How can I calculate the following integral : 
$$\int\frac{1}{(x+1)\sqrt{1+x^2}} dx $$
I try to write the integral like :
$$\int\frac{1+x-x}{(x+1)\sqrt{1+x^2}}=\int\frac{1}{\sqrt{1+x^2}}-\int\frac{x}{(x+1)\sqrt{1+x^2}}=\int\frac{1}{\sqrt{1+x^2}}-\int\frac{(\sqrt{x^2+1})'}{(x+1)}$$
but still nothing .
thanks :) 
 A: Let $x=\tan[y]$, then $dx=\sec^{2}[y]dy$ $$\frac{1}{1+x}=\frac{\cos[y]}{\sin[y]+\cos[y]},\frac{1}{\sqrt{1+x^{2}}}=\frac{1}{\sec[y]}=\cos[y]$$
Multiplying everything out you need to integrate $$\frac{1}{\sin[y]+\cos[y]}dy$$
But you can simplify $$\sin[y]+\cos[y]=\sqrt{2}[\sin[y]\cos[\frac{\pi}{4}]+\cos[y]\sin[\frac{\pi}{4}]]=\sqrt{2}\sin[y+\frac{\pi}{4}]$$
So it suffice to integrate $$\frac{1}{\sqrt{2}}\csc[y+\frac{\pi}{4}]dy$$And we know how to do $\int \csc[x]dx$. 
A: Try to substitute $x=\tan t$ and after some manipulations recall that
$$
\left(\ln \tan \frac{u}{2}\right)'= \frac{u'}{\sin u}
$$
A: Substitute $x=\frac{1-t}{1+t}$
$$\int\frac{1}{(x+1)\sqrt{1+x^2}} dx =-\int\frac1{\sqrt{2(1+t^2)}}dt=-\frac1{\sqrt2}\sinh^{-1}t
$$
A: This is how I would go, term $\sqrt{1+x^2}$ makes me go for $x=\sinh(u)$, which will convert to $\cosh(u)$ and cancel with the one brought by $du$.
Then I convert to exponential form ($t=e^u$) to get a rational fraction.
$\begin{align}\require{cancel}\int\frac {dx}{(x+1)\sqrt{1+x^2}}
&=\int\frac {\cancel{\cosh(u)}du}{(\sinh(u)+1)\cancel{\sqrt{1+\sinh(u)^2}}}
=\int\frac {du}{\sinh(u)+1}\\\\
&=\int\frac {dt/t}{(t-1/t)/2+1}
=\int\frac {2\,dt}{t^2+2t-1}
=\int\frac {2\,dt}{(t+1)^2-2}\\\\
&=C-\sqrt{2}\tanh^{-1}\Big(\tfrac{t+1}{\sqrt{2}}\Big)\end{align}$
The substitutions proposed appear quite naturally, though the final result (especially after back substitution) is not as appealing as the one proposed by Quanto...
