How to integrate $\int\frac{\sqrt{1+x^2}}{x}\,\mathrm dx$

Question

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

Could anyone solve it?

Thanks

Answer 1

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

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

Answer 2

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

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

$$\int\frac{\sqrt{1+x^2}}{x}\,dx=\int\frac{v^2}{v^4-1}\,dv$$

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.

Answer 3

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.

Answer 4

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

Answer 5

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