How to integrate $\int\frac{\sqrt{1+x^2}}{x}\,\mathrm dx$ I want to know how to integrate 
$$\int\frac{\sqrt{1+x^2}}{x}\,\mathrm dx$$
Could anyone solve it?
Thanks
 A: Let $x = \tan \theta \implies dx = \sec^2 \theta d\theta$
$$\begin{align} \int \frac{\sqrt{ 1 + x^2}}{x}\,dx & = \int\frac{d\theta}{\sin\theta\cos^2\theta}\\ \\ & = \int \csc \theta \sec^2\theta \,d\theta \\ \\ &= \int \csc \theta(1 + \tan^2\theta)\,d\theta \\ \\ & = \int \csc\theta \,d\theta + \int \csc\theta\tan^2 \theta \,d\theta \\ \\ & = \int \csc\theta \,d\theta + \int \dfrac{\sin\theta}{\cos^2 \theta}\,d\theta\end{align}$$
A: Working backwards from an answer by WolframAlpha one obtains the following.
We set $u=\sqrt{x^2+1}$. Then $u^2=x^2+1$ and $2u\,du=2x\,dx$. Being careful not to
forget that $u$ and $x$ are not independent variables we calculate
\begin{align*}
\int\frac {\sqrt{x^2+1}}x\,dx&=
\int\frac ux\,dx=\int\frac{u-1}x\,dx+\int\frac1x\,dx=
\int\frac{(u-1)u}{x^2}\,du+\int\frac1x\,dx
=\\&=
\int\frac{(u-1)u}{u^2-1}\,du+\int\frac1x\,dx
=
\int\frac{u}{u+1}\,du+\int\frac1x\,dx
=\\&=
\int 1-\frac{1}{u+1}\,du+\int\frac1x\,dx
=\\&=u-\ln(u+1)+\ln x +C 
=\\&=\sqrt{x^2+1}-\ln\left(\sqrt{x^2+1}+1\right)+\ln x +C.
\end{align*}
This is admittedly not the most elegant solution.
I had meant to append the following to user64494's answer, but it was rejected as too big a change. Fair enough.
If one looks at Maple's solution one sees the two key steps: Substitute $u=\sqrt{x^2+1}$, use partial fraction decomposition afterwards. Using this one easily arrives at the following.
Differentiating $u^2=x^2+1$ yields $2u\,du=2x\,dx$ and hence
\begin{align*}
\int\frac{\sqrt{x^2+1}}x\,dx &=
\int\frac ux\,dx =
\int \frac{u^2}{x^2}\,du=\int\frac{u^2}{u^2-1}\,du
=\int1+\frac{1/2}{u-1}+\frac{-1/2}{u+1}\,du
\\&= u + \frac12\ln(u-1)-\frac12\ln(u+1) + C.
\end{align*}
Note that some people will frown at the integrals containing both $x$ and $u$, so one might want to avoid these.
A: $$\int\frac{\sqrt{1+x^2}}{x}\,dx=\int\frac{x\sqrt{1+x^2}}{x^2}\,dx$$
Let $u=x^2+1, du =2x dx$. Then
$$\int\frac{\sqrt{1+x^2}}{x}\,dx= \frac{1}{2}\int\frac{\sqrt{u}}{u^2-1}\,du$$
If $v=\sqrt{u}$ the n $u=v^2, du=2vdv$. Thus
$$\int\frac{\sqrt{1+x^2}}{x}\,dx=\int\frac{v^2}{v^4-1}\,dv$$
By Partial Fraction Decomposition 
$$\frac{v^2}{v^4-1}=\frac{A}{v-1}+\frac{B}{v+1}+\frac{Cv+D}{v^2+1}$$
find $A,B,C,D$, and use $v=\sqrt{u}=\sqrt{x^2+1}$ and you are done.
A: You can change the form of the function to: $$\frac{1+x^2}{x^2} \frac{x}{\sqrt{1+x^2}}$$.
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
&\color{#0000ff}{\large\int{\root{1 + x^{2}} \over x}\,\dd x}
= \half\int{\root{1 + x^{2}} \over x^{2}}\,\dd\pars{x^{2}}
= \half\overbrace{\int{\root{1 + y} \over y}\,\dd y}^{y = x^{2}}
=\half\overbrace{\int{z \over z^{2} - 1}\,2z\dd z}^{z = \root{1 + y}}
\\[3mm]&=\int\bracks{1 + \half\,\pars{{1 \over z - 1} - {1 \over z + 1}}}\,\dd z
=
z + \half\,\ln\pars{z - 1 \over z + 1}
=
\root{1 + y} + \half\,\ln\pars{\root{1 + y} - 1 \over \root{1 + y} + 1}
\\[3mm]&
=
\color{#0000ff}{\large\root{1 + x^{2}} + \half\,\ln\pars{\root{1 + x^{2}} - 1 \over \root{1 + x^{2}} + 1}} + \mbox{a constant}.
\end{align}
