Integrate $\int\frac{dx}{x\sqrt{x^2+1}}$ I would like to ask for some help regarding the following indefinite integral, tried integration by parts and trigonometric substitution which both brought me to $\int\frac{\sec\theta}{\tan\theta}d\theta$, and from this point it is messy to integrate by parts, any help would be appreciated.
$$\int\frac{dx}{x\sqrt{x^2+1}}$$
 A: Multiply the integrand by $\dfrac{x}{x}$, we will have
$$
\int\frac{x\ dx}{x^2\sqrt{x^2+1}}\ dx.
$$
Now, set $u^2=x^2+1\ \Rightarrow\ u\ du=x\ dx$ then
\begin{align}
\int\frac{x\ dx}{x^2\sqrt{x^2+1}}\ dx&=\int\frac{1}{u^2-1}\ du\\
&=\frac12\int\left[\frac1{u-1}-\frac1{u+1}\right]\ du.
\end{align}
The rest should be easy.
A: $$\int \frac{\sec \theta}{\tan\theta} = \frac{\frac 1{\cos \theta}}{\frac {\sin\theta}{\cos\theta}}\,d\theta = \int \frac 1{\sin\theta}\,d\theta = \int \csc\theta \,d\theta$$
Alternatively, given $$\int\frac{dx}{x\sqrt{x^2+1}} = \int\frac{x\,dx}{x^2 \sqrt{x^2 + 1}}$$
$$\text{Put }\;x^2 + 1 = u^2\;\iff \;x^2 = u^2 - 1\; \implies \;u\,du = x\,dx$$
This gives us the integral, after substitution: $$\int \frac{u\,du}{(u^2-1)u}=\int \frac{du}{(u^2-1)} = \frac 12\int \left(\frac 1{u-1} - \frac 1{u+1}\right)\,du$$
I'm sure you can take it from here.
A: Even another aproach to the integral (albeit rather in "philosophical" form in some sence); by substitution $x=1/t$ :
$$\int\frac{\mathrm{d}x}{x\sqrt{1+x^2}}=-\int\frac{\mathrm{d}t}{\sqrt{1+t^2}} = -\operatorname{arcsinh}t = -\operatorname{arcsinh}\frac{1}{x} = -\operatorname{arccsch}x + C $$
A: $$
\begin{aligned}
\int \frac{d x}{x \sqrt{x^{2}+1}} =& \int \frac{1}{x^{2}} d\left(\sqrt{x^{2}+1}\right) \\
=& \int \frac{1}{\left(\sqrt{x^{2}+1}\right)^{2}-1} d (\sqrt{x^{2}+1} )\\
=& \frac{1}{2} \ln \left|\frac{\sqrt{x^{2}+1}-1}{\sqrt{x^{2}+1}+1}\right|+C
\end{aligned}
$$
A: Let $x=\sinh(y),$ hence $$\int\frac1{x\sqrt{x^2+1}}\,\mathrm{d}x=\int\frac1{\sinh(y)\sqrt{\cosh{y}^2}}\cosh(y)\,\mathrm{d}y=\int\frac1{\sinh(y)}\,\mathrm{d}y=\int\frac2{e^y-e^{-y}}\,\mathrm{d}y=\int\frac{2e^y}{e^{2y}-1}\,\mathrm{d}y.$$ Let $z=e^y=e^{\operatorname{arsinh}(x)}=x+\sqrt{x^2+1},$ hence $$\int\frac{2e^y}{e^{2y}-1}\,\mathrm{d}y=\int\frac2{z^2-1}\,\mathrm{d}z=\int\frac{(z+1)-(z-1)}{z^2-1}\,\mathrm{d}z=\int\frac1{z-1}-\frac1{z+1}\,\mathrm{d}z=\begin{cases}\ln\left(-\frac{z-1}{z+1}\right)+C_0&|z|\lt1\\\ln\left(\frac{z-1}{z+1}\right)+C_1&|z|\gt1\end{cases}=\begin{cases}\ln\left(-\frac{x+\sqrt{x^2+1}-1}{x+\sqrt{x^2+1}+1}\right)+C_0&x\lt0\\\ln\left(\frac{x+\sqrt{x^2+1}-1}{x+\sqrt{x^2+1}+1}\right)+C_1&x\gt0\end{cases}$$
A: $\displaystyle\int\frac{dx}{x\sqrt{x^{2}+1}}$
$\displaystyle z^{2}=x^{2}+1\Rightarrow 2zdz=2xdx\Rightarrow xzdz=x^{2}dx=
(z^{2}-1)dx$
$\displaystyle\Rightarrow \frac{dz}{z^{2}-1}=\frac{dx}{xz}$
$\displaystyle\int\frac{dx}{x\sqrt{x^{2}+1}}=\int\frac{dz}{z^{2}-1}=\frac{1}{2}\int(\frac{1}{z-1}-\frac{1}{z+1})dz$
$\displaystyle\int\frac{dx}{x\sqrt{x^{2}+1}}=\frac{1}{2}ln\left| \frac{\sqrt{x^{2}+1}-1}{\sqrt{x^{2}+1}+1} \right|+c$
