How to integrate $\frac{1}{x\sqrt{1+x^2}}$ using substitution? How you integrate 
$$\frac{1}{x\sqrt{1+x^2}}$$
using following substitution? $u=\sqrt{1+x^2} \implies du=\dfrac{x}{\sqrt{1+x^2}}\, dx$ 
And now I don't know how to proceed using substitution rule.
 A: If $u=\sqrt{1+x^2}$ then $u^2 = 1+x^2$, so $x^2= u^2-1$.  Then you have
\begin{align}
& \int \frac 1 {x\sqrt{1+x^2}} \,dx = \int \frac {x} {x^2\sqrt{1+x^2}} \,dx \\[8pt]
= {} & \int\frac{du}{u^2-1} = \int\frac{du}{(u-1)(u+1)}.
\end{align}
Then use partial fractions.
A: Use $x=\tan\theta$, $dx=\sec^2\theta\,d\theta$
$\tan^2\theta+1=\sec^2\theta$
$$\int\dfrac{\sec^2\theta\,d\theta}{\tan\theta\sec\theta}=\int\dfrac{\sec\theta\,d\theta}{\tan\theta}=\int\dfrac{d\theta}{\sin\theta}=-\ln|\csc\theta+\cot\theta|+C$$
A: Let us consider $$I=\int \frac{dx}{x\sqrt{1+x^2}}$$ and make $u=\sqrt{1+x^2}$ that is to say $x=\sqrt{u^2-1}$ and then $dx=\frac{u}{\sqrt{u^2-1}}du$. So, $$I=\int \frac{du}{u^2-1}$$ Now, use partial fraction decomposition and you are done.
A: Let $x=\tan\theta$, then $dx=\sec^2\theta\,d\theta$, now
$$\int{dx\over x\sqrt{1+x^2}}=\int{\sec\theta\,d\theta\over\tan\theta}=\int\csc\theta\,d\theta.$$
Can you take it from here?
A: Maybe another substitution is interesting:
For $x>0, v=\frac{1}{x}$
$$
\int \frac 1 {x\sqrt{1+x^2}} \,dx = \int \frac {1} {x^2\sqrt{\frac{1}{x^2}+1}} \,dx = -\int\frac{dv}{\sqrt{v^2+1}} = - \ln(v+\sqrt{v^2+1})=$$
$$ =- \ln\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}+1}\right) + C.
$$
For $x<0, v=\frac{1}{x}$
$$
\int \frac 1 {x\sqrt{1+x^2}} \,dx = -\int \frac {1} {x^2\sqrt{\frac{1}{x^2}+1}} \,dx = \int\frac{dv}{\sqrt{v^2+1}} =  \ln(v+\sqrt{v^2+1})+ C=$$
$$ = \ln\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}+1}\right) + C.
$$
A: It is quickest to substitute only in part of the integral. The substitution $u=\sqrt{1+x^2}$ satisfies
$$
\frac{1}{xu}=\frac{1}{x}-\frac{1}{1+u}\frac{x}{u} \quad \text{and} \quad \mathrm{d}u=\frac{x}{u}\,\mathrm{d}x.
$$
So it follows that
$$
\int \frac{\mathrm{d}x}{x\sqrt{1+x^2}}=\int \frac{\mathrm{d}x}{x}-\int \frac{\mathrm{d}u}{1+u}=\log \frac{|x|}{1+u}+\mathrm{const}=-\operatorname{arsinh}\frac{1}{|x|}+\mathrm{const}.
$$
Here we have used the following identity for the inverse hyperbolic sine:
$$
\operatorname{arsinh}t=\log\,(t+\sqrt{t^2+1}).
$$
A: $$\int\frac{dx}{x\sqrt{x^2+1}}~=~\int\frac x{x^2\sqrt{x^2+1}}~dx~=~\frac12\int\frac{d\big(x^2+1\big)}{\Big[(x^2+1)-1\Big]\sqrt{x^2+1}}~=~\frac12\int\frac{dt}{(t-1)\sqrt t}$$
Now let $t=u^2$, and use partial fraction decomposition. $($Alternately, let $x=\sinh y)$.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
With the substitution
$\ds{x \equiv \frac{1 - t^{2}}{2t}\ \imp\ t=\root{1 + x^{2}} - x}$:
\begin{align}
\color{#66f}{\int\frac{\dd x}{x\root{1 + x^{2}}}}
&=\int\pars{\frac{1}{t - 1} - \frac{1}{t + 1}}\,\dd t
=\ln\pars{\verts{\frac{t - 1}{t + 1}}}
\\[5mm]&=\color{#66f}{%
\ln\pars{\verts{\frac{\root{1 + x^{2}} - x - 1}{\root{1 + x^{2}} - x + 1}}}}
+ \mbox{a constant}
\end{align}
