Evaluate the integral $\int \:\frac{\sqrt{x^2+2\cdot\:\:x+2}}{x}dx$ $$\int \:\frac{\sqrt{x^2+2\cdot\:\:x+2}}{x}dx$$
I have tried to form a square above i also tried to get the x below under the root
but got nothing
 A: The standard way is to make the substitution
$y=x+\sqrt{x^2+2x+2}$
so that $x=\frac{y^2-2}{2(y+1)}$
$\sqrt{x^2+2x+2}=\frac{y^2+2y+2}{2(y+1)}$
$dx=\frac{y^2+2y+2}{2(y+1)^2}dy$
which changes your integral into an integral of rational function, solvable by partial fraction expansion:
$\int\frac{(y^2+2y+2)^2}{2(y^2-2)(y+1)^2}dy$
It is messy to solve it, but it is definitely doable with the right ammount of determination.
The partial fraction expansion of $\frac{(y^2+2y+2)^2}{2(y^2-2)(y+1)^2}$ is:
$\frac{4}{y^2-2}+\frac{1}{y+1}-\frac{1}{2(y+1)^2}+\frac{1}{2}$
which you integrate term by term.
There might be a easier trick to solve it, but i don't see it.

EDIT: after more thinking i think that one can make the following substitution
$x+1=\tan\theta$ with $\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$
$\sqrt{x^2+2x+2}=\sqrt{(\tan\theta)^2+1}=\sqrt\frac{1}{\cos^2\theta}=\frac{1}{\cos\theta}$
as $\cos\theta>0$ when $\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$
Then $dx=\frac{1}{\cos^2\theta} d\theta$
So finally you integrate:
$\int \frac{1}{\cos^2\theta(\sin\theta-\cos\theta)}d\theta$, with $\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$
From here one might go to the tangent of half angle, but again I see no easy formula.
A: Rationalise the numerator.
$\displaystyle \int \frac{\sqrt{x^2+2x+2}}{x} \, \text{d}x = \int \frac{x^2+2x+2}{x\sqrt{x^2+2x+2}}\, \text{d}x$
Split apart as follows:
$\displaystyle \int \frac{x^2+2x+2}{x\sqrt{x^2+2x+2}}\, \text{d}x = \int \frac{x+1}{\sqrt{x^2+2x+2}} \, \text{d}x + \int \frac{\text{d}x}{\sqrt{x^2+2x+2}} + \int \frac{2\,\text{d}x}{x\sqrt{x^2+2x+2}}$
The first is an application of the reverse chain rule, the second is an application of the inverse hyperbolic sine formula, and the third requires the substitution $\displaystyle u = \frac{2}{x}$.
$\displaystyle \int \frac{\sqrt{x^2+2x+2}}{x} \, \text{d}x = \sqrt{x^2+2x+2} + \log{\left(x+1+\sqrt{x^2+2x+2} \right)} - \int \frac{\frac{4 \text{d}u}{u^2}}{\frac{2}{u} \sqrt{\frac{4}{u^2}+\frac{4}{u}+2}} $
$\displaystyle \int \frac{\frac{4 \text{d}u}{u^2}}{\frac{2}{u} \sqrt{\frac{4}{u^2}+\frac{4}{u}+2}} = \int \frac{\sqrt{2}\,\text{d}u}{\sqrt{u^2+2u+2}} = \sqrt{2} \log{\left( u+1+\sqrt{u^2+2u+2} \right)}$
$\displaystyle \int \frac{\sqrt{x^2+2x+2}}{x} \, \text{d}x = \sqrt{x^2+2x+2} + \log{\left(x+1+\sqrt{x^2+2x+2} \right)} - \sqrt{2} \log{\left( \frac{2}{x}+1+\sqrt{\frac{4}{x^2}+\frac{4}{x}+2} \right)} $
$\displaystyle = \sqrt{x^2+2x+2} + \log{\left(x+1+\sqrt{x^2+2x+2} \right)} +\sqrt{2}\log{x} - \sqrt{2}\log{\left( x+2+\sqrt{2}\sqrt{x^2+2x+2} \right)}$
A: Continuing from Momo's answer where was proposed the interesting change of variable $x=\tan(\theta)-1$, the integral reduces to $$I=\int \frac{d\theta}{\cos^2(\theta)\,(\sin(\theta)-\cos(\theta))}$$ Using the tangent half-angle substitution (just as Momo proposed), it reduces to $$I=\int\frac{2 \left(t^2+1\right)^2}{\left(t^2-1\right)^2 (t^2+2t-1)}\,dt$$ Using partial fraction decomposition $$\frac{2 \left(t^2+1\right)^2}{\left(t^2-1\right)^2 (t^2+2t-1)}=\frac{4}{t^2+2 t-1}+\frac{1}{t+1}-\frac{1}{(t+1)^2}-\frac{1}{t-1}+\frac{1}{(t-1)^2}$$ which makes $$I=\frac 2{1-t^2}+\log\left(\frac{1+t}{1-t}\right)+\sqrt 2\log\left(\frac{\sqrt 2-1-t}{\sqrt 2+1+t}\right)$$
A: $z=\tan(\theta/2)$, $\tan(\theta)=\frac{2z}{1-z^2}$, $\mathrm{d}\theta=\frac{2\,\mathrm{d}z}{1+z^2}$, $\cos(\theta)=\frac{1-z^2}{1+z^2}$, $\sin(\theta)=\frac{2z}{1+z^2}$
$z=\frac{\sec(\theta)-1}{\tan(\theta)}=\frac{-1+\sqrt{x^2+2x+2}}{x+1}$
$$
\begin{align}
\int\frac{\sqrt{x^2+2x+2}}{x}\,\mathrm{d}x
&=\int\frac{\sec^3(\theta)}{\tan(\theta)-1}\,\mathrm{d}\theta\\
&=\int\left(\frac{1+z^2}{1-z^2}\right)^3\frac{1-z^2}{z^2+2z-1}\frac{2\,\mathrm{d}z}{1+z^2}\\
&=\int\left(\frac{1+z^2}{1-z^2}\right)^2\frac{2\,\mathrm{d}z}{z^2+2z-1}\\
&=\int\frac{2z^4+4z^2+2}{(z+1+\sqrt2)(z+1-\sqrt2)(z-1)^2(z+1)^2}\,\mathrm{d}z\\
&=\int\left(\frac1{(z-1)^2}-\frac1{z-1}-\frac1{(z+1)^2}+\frac1{z+1}+\frac{\sqrt2}{z+1-\sqrt2}-\frac{\sqrt2}{z+1+\sqrt2}\right)\mathrm{d}z\\
&=-\frac2{z^2-1}+\log\left(\frac{z+1}{z-1}\right)+
\sqrt2\log\left(\frac{z+1-\sqrt2}{z+1+\sqrt2}\right)+C
\end{align}
$$
Now back substitute for $x$.
