How to integrate $\frac{1}{x\sqrt{1+x^2}}$ using substitution?

Question

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.

Answer 1

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.

Answer 2

Use $x=\tan\theta$, $dx=\sec^2\theta\,d\theta$

$\tan^2\theta+1=\sec^2\theta$

$$\int\dfrac{\sec^2\theta\,d\theta}{\tan\theta\sec\theta}=\int\dfrac{\sec\theta\,d\theta}{\tan\theta}=\int\dfrac{d\theta}{\sin\theta}=-\ln|\csc\theta+\cot\theta|+C$$

Answer 3

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.

Answer 4

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?

Answer 5

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 $x<0, v=\frac{1}{x}$ $$ \int \frac 1 {x\sqrt{1+x^2}} \,dx = -\int \frac {1} {x^2\sqrt{\frac{1}{x^2}+1}} \,dx = \int\frac{dv}{\sqrt{v^2+1}} = \ln(v+\sqrt{v^2+1})+ C=$$ $$ = \ln\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}+1}\right) + C. $$

Answer 6

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:

$$ \operatorname{arsinh}t=\log\,(t+\sqrt{t^2+1}). $$

Answer 7

$$\int\frac{dx}{x\sqrt{x^2+1}}~=~\int\frac x{x^2\sqrt{x^2+1}}~dx~=~\frac12\int\frac{d\big(x^2+1\big)}{\Big[(x^2+1)-1\Big]\sqrt{x^2+1}}~=~\frac12\int\frac{dt}{(t-1)\sqrt t}$$

Now let $t=u^2$, and use partial fraction decomposition. $($Alternately, let $x=\sinh y)$.

Answer 8

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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}