Integrate $\int\frac{dx}{x\sqrt{x^2+1}}$

Question

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.

$$\int\frac{dx}{x\sqrt{x^2+1}}$$

Answer 1

$$\int \frac{\sec \theta}{\tan\theta} = \frac{\frac 1{\cos \theta}}{\frac {\sin\theta}{\cos\theta}}\,d\theta = \int \frac 1{\sin\theta}\,d\theta = \int \csc\theta \,d\theta$$

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.

Answer 2

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.

Answer 3

Even another aproach to the integral (albeit rather in "philosophical" form in some sence); by substitution $x=1/t$ :

$$\int\frac{\mathrm{d}x}{x\sqrt{1+x^2}}=-\int\frac{\mathrm{d}t}{\sqrt{1+t^2}} = -\operatorname{arcsinh}t = -\operatorname{arcsinh}\frac{1}{x} = -\operatorname{arccsch}x + C $$

Answer 4

$$ \begin{aligned} \int \frac{d x}{x \sqrt{x^{2}+1}} =& \int \frac{1}{x^{2}} d\left(\sqrt{x^{2}+1}\right) \\ =& \int \frac{1}{\left(\sqrt{x^{2}+1}\right)^{2}-1} d (\sqrt{x^{2}+1} )\\ =& \frac{1}{2} \ln \left|\frac{\sqrt{x^{2}+1}-1}{\sqrt{x^{2}+1}+1}\right|+C \end{aligned} $$

Answer 5

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 $$\int\frac{2e^y}{e^{2y}-1}\,\mathrm{d}y=\int\frac2{z^2-1}\,\mathrm{d}z=\int\frac{(z+1)-(z-1)}{z^2-1}\,\mathrm{d}z=\int\frac1{z-1}-\frac1{z+1}\,\mathrm{d}z=\begin{cases}\ln\left(-\frac{z-1}{z+1}\right)+C_0&|z|\lt1\\\ln\left(\frac{z-1}{z+1}\right)+C_1&|z|\gt1\end{cases}=\begin{cases}\ln\left(-\frac{x+\sqrt{x^2+1}-1}{x+\sqrt{x^2+1}+1}\right)+C_0&x\lt0\\\ln\left(\frac{x+\sqrt{x^2+1}-1}{x+\sqrt{x^2+1}+1}\right)+C_1&x\gt0\end{cases}$$

Answer 6

$\displaystyle\int\frac{dx}{x\sqrt{x^{2}+1}}$

$\displaystyle z^{2}=x^{2}+1\Rightarrow 2zdz=2xdx\Rightarrow xzdz=x^{2}dx= (z^{2}-1)dx$

$\displaystyle\Rightarrow \frac{dz}{z^{2}-1}=\frac{dx}{xz}$

$\displaystyle\int\frac{dx}{x\sqrt{x^{2}+1}}=\int\frac{dz}{z^{2}-1}=\frac{1}{2}\int(\frac{1}{z-1}-\frac{1}{z+1})dz$

$\displaystyle\int\frac{dx}{x\sqrt{x^{2}+1}}=\frac{1}{2}ln\left| \frac{\sqrt{x^{2}+1}-1}{\sqrt{x^{2}+1}+1} \right|+c$

Answer 7

Since this is an integral of the form $\int\frac{\mathrm dx}{L\sqrt Q}$, a feasible substitution is $x=\frac1t$:

$$\begin{align}\int\frac{\mathrm dx}{x\sqrt{x^2+1}}&=-\text{sgn}(x)\int\frac{\mathrm dt}{\sqrt{1+t^2}}\\&=-\text{sgn}(x)\sinh^{-1}t+C\\&=-\text{sgn}(x)\text{csch}^{-1}x+C\end{align}$$