Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to know how to integrate $$\int\frac{\sqrt{1+x^2}}{x}\,dx$$

Could anyone solve it?


share|improve this question
What have you tried? –  Ahaan Rungta Dec 3 '13 at 15:28
Hint: Substitute $ x = \tan \phi $. –  Ahaan Rungta Dec 3 '13 at 15:29
Use the substion $\sqrt{1+x^2}=t-x$. Now we have $x=\frac{t^2-1}{2t},$ $\sqrt{1+x^2}=\frac{t^2+1}{2t},$, and $dx=\frac{t^2+1}{2t^2}dt,$ and for the gven integral now we have $$\int\frac{\sqrt{1+x^2}}{x}=\int\frac{(t^2+1)^2 dt}{2t^2(t^2-1)}$$ –  Madrit Zhaku Dec 3 '13 at 15:46
@yes I have tried put x =tan(&) the it becomes sec^3(&)/tan(&) the it becomes 1/(sin(&)cos^2(&) so? –  user32104 Dec 3 '13 at 15:47
in the case $x=\tan(\phi)$ we have $$\int\frac{d\phi}{\sin\phi\cos^2\phi}$$ –  Madrit Zhaku Dec 3 '13 at 15:49

6 Answers 6

up vote 3 down vote accepted

Let $x = \tan \theta \implies dx = \sec^2 \theta d\theta$

$$\begin{align} \int \frac{\sqrt{ 1 + x^2}}{x}\,dx & = \int\frac{d\theta}{\sin\theta\cos^2\theta}\\ \\ & = \int \csc \theta \sec^2\theta \,d\theta \\ \\ &= \int \csc \theta(1 + \tan^2\theta)\,d\theta \\ \\ & = \int \csc\theta \,d\theta + \int \csc\theta\tan^2 \theta \,d\theta \\ \\ & = \int \csc\theta \,d\theta + \int \dfrac{\sin\theta}{\cos^2 \theta}\,d\theta\end{align}$$

share|improve this answer
so what is the integral of these ? –  user32104 Dec 3 '13 at 15:59
For the second, use the substitution $u = \cos \theta \implies du = -\sin\theta \,d\theta$ to get $$\int -\dfrac{du}{u^2} = \dfrac 1u + C$$ –  amWhy Dec 3 '13 at 16:06
But $\int \csc \theta \, d\theta = -\ln \left( \csc \left( \theta \right) +\cot \left( \theta \right) \right) +C $. –  user64494 Dec 3 '13 at 16:10
@user64494 Thank you! Yes, I was working on the two integrals simultaneously and got mixed up on the first part of my comment. –  amWhy Dec 3 '13 at 16:13
See the original comment by amWhy here. –  user64494 Dec 3 '13 at 16:15

You can change the form of the function to: $$\frac{1+x^2}{x^2} \frac{x}{\sqrt{1+x^2}}$$.

share|improve this answer

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.

share|improve this answer

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}

share|improve this answer


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


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.

share|improve this answer

The Maple command $$Student[Calculus1]:-IntTutor(sqrt(x^2+1)/x, x) $$ does it step by step with explanation, producing $$\sqrt {{x}^{2}+1}+1/2\,\ln \left( \sqrt {{x}^{2}+1}-1 \right) -1/2\, \ln \left( \sqrt {{x}^{2}+1}+1 \right) .$$ See that link and the output for info.

share|improve this answer
The downvoters: Don't laugh the people. –  user64494 Dec 3 '13 at 16:02
I have seen your 'work' here user. Have you yet been accused of being a rep for Maple? I am just wondering if it has arisen. Do I think you are? No I don't. –  Jp McCarthy Dec 3 '13 at 16:06
Maple does it better. –  user64494 Dec 3 '13 at 16:07
That isn't what I asked you. –  Jp McCarthy Dec 3 '13 at 16:08
And I had wanted to answer wolframalpha.com/input/… –  Carsten Schultz Dec 3 '13 at 16:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.