Evaluating $\int{\frac{1}{\sqrt{x^2-1}(x^2+1)}dx}$

Question

Evaluating $$\int{\frac{1}{\sqrt{x^2-1}(x^2+1)}\mathrm dx}$$ using $ux=\sqrt{x^2-1}$
UPDATE 'official' solution

$$u^2x^2=x^2-1$$
$$x^2=\frac{-1}{u^2-1}$$ $$x^2+1=\frac{u^2-2}{u^2-1}$$ $$2xdx=\frac{-2u}{(u^2-1)^2}$$

So, $$\int{\frac{x}{x\sqrt{x^2-1}(x^2+1)}\mathrm dx}$$ $$=\int{\frac{1}{\frac{-1}{u^2-1}u\frac{u^2-2}{u^2-1}}\left(\frac{-u}{(u^2-1)^2}\right)\mathrm du} $$ $$=\int{\frac{1}{u^2-2}\mathrm du} $$ $$=\frac{\log \left|\sqrt{2}-x\right|-\log \left|x+\sqrt{2}\right|}{2 \sqrt{2}}+C$$ $$=\frac{\log \left|\sqrt{2}-\frac{\sqrt{x^2-1}}x\right|-\log \left|\frac{\sqrt{x^2-1}}x+\sqrt{2}\right|}{2 \sqrt{2}}+C$$ However at mathematica I get $$\frac{\log \left(3 x^2-2 \sqrt{2} \sqrt{x^2-1} x+1\right)-\log \left(-3 x^2+2 \sqrt{2} \sqrt{x^2-1} x+1\right)}{4 \sqrt{2}}+C$$ Where is the error?

Answer 1

$$u=\frac{\sqrt{x^2-1}}x$$

$$\frac{du}{dx}=-\frac{\sqrt{x^2-1}}{x^2}+\frac{2x}{x\cdot2\sqrt{x^2-1}}$$

$$=\frac{-(x^2-1)+x^2}{x^2\sqrt{x^2-1}}$$

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

Now $u^2=\dfrac{x^2-1}{x^2}\implies\dfrac1{x^2}=1-u^2$

$\implies\dfrac{x^2}{1+x^2}=\dfrac1{1+1/x^2}=\dfrac1{1+1-u^2}$

Answer 2

Let us evaluate the general form of the integral \begin{align} \int\frac{\mathrm dx}{(x^2+a^2)\sqrt{x^2-b^2}}&=\int\frac{a\sec^2t}{(a^2\tan^2t+a^2)\sqrt{a^2\tan^2t-b^2}}\mathrm dt\tag1\\[7pt] &=\frac{1}{a}\int\frac{\cos t}{\sqrt{a^2\sin^2t-b^2\cos^2t}}\mathrm dt\tag2\\[7pt] &=\frac{1}{a}\int\frac{\cos t}{\sqrt{\left(a^2+b^2\right)\sin^2t-b^2}}\mathrm dt\tag3\\[7pt] &=\frac{1}{a\sqrt{a^2+b^2}}\int\frac{\mathrm dy}{\sqrt{y^2-1}}\tag4\\[7pt] &=\frac{\operatorname{arccosh} y}{a\sqrt{a^2+b^2}}+C\tag5\\[7pt] &=\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\frac{1}{a\sqrt{a^2+b^2}}\operatorname{arccosh}\left(\frac{x\,\sqrt{a^2+b^2}}{b\left(x^2+a^2\right)}\right)+C}} \end{align} Setting $a=1$ and $b=1$, we get $$\int\frac{\mathrm dx}{(x^2+1)\sqrt{x^2-1}} =\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\frac{1}{\sqrt{2}}\operatorname{arccosh}\left(\frac{x\,\sqrt{2}}{x^2+1}\right)+C}}$$

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Explanation :

$(1)\;$ Use substitution $\;\displaystyle x=a\tan t\quad\implies\quad\tan t=\frac{x}{a}$

$(2)\;$ Use identities $\;\displaystyle \sec^2t=1+\tan^2t\;$ and $\;\displaystyle \tan t=\frac{\sin t}{\cos t}$

$(3)\;$ Use identity $\;\displaystyle \cos^2t=1-\sin^2t$

$(4)\;$ Use substitution $\;\displaystyle \sin t=\frac{by}{\sqrt{a^2+b^2}}\,$

$(5)\;$ Use substitution $\;\displaystyle y=\cosh u\,$. We have $\;\displaystyle y=\frac{\sqrt{a^2+b^2}\sin t}{b}\,$ and $\;\displaystyle \sin t=\frac{x}{\sqrt{x^2+a^2}}\,$

Answer 3

let me try a change of variable $$x = {1 + t^2 \over 1 - t^2},\ dx = {-4t \ dt \over (1-t^2)^2},\ x^2+1 = {2(1+t^4) \over (1-t^2)^2}, \ x^2-1 = {4t^2 \over (1-t^2)^2}$$

$$\int {1 \over (x^2+1) \sqrt{x^2-1}} dx = \int {t^2 - 1 \over t^4 + 1} \ dt = \int {1 \over t^2 + 1} dt - 2 \int {1 \over t^4 + 1} dt = \tan^{-1}t - 2\int{dt \over t^4+1}$$

i got to go now. i will see if can do the last integral later.

edit: Idris, thanks for the hint. i can now evaluate

$ \begin{eqnarray} 2\int {dt \over 1 + t^4} &=& 2\int {1/t^2 \over t^2 + 1/t^2}dt = \int {1+1/t^2 \over t^2 + 1/t^2} \ dt - \int {1-1/t^2 \over t^2 + 1/t^2} \ dt \\ &=&\int {d(t-1/t)\over (t-1/t)^2 + 2} - \int {d(t+1/t)\over (t+1/t)^2 - 2} \\ &=&{1 \over \sqrt 2}\tan^{-1}\left({t - 1/t \over \sqrt 2} \right) - {1 \over 2\sqrt 2}\ln\left({t + 1/t -\sqrt2 \over t + 1/t +\sqrt 2} \right) \end{eqnarray} $

so finally, $$\int {1 \over (x^2+1) \sqrt{x^2-1}} dx = \tan^{-1}t - {1 \over \sqrt 2}\tan^{-1}\left({t - 1/t \over \sqrt 2} \right) + {1 \over 2\sqrt 2}\ln\left({t + 1/t -\sqrt2 \over t + 1/t +\sqrt 2} \right) + C$$

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second edit: sorry, i made a mistake and i will fix it.

$\begin{eqnarray} \int {1 \over (x^2+1) \sqrt{x^2-1}} dx &=& \int {t^2 - 1 \over t^4 + 1} \ dt =\int{1 - 1/t^2 \over t^2 + 1/t^2}\ dt = \int {d(t+1/t) \over (t+1/t)^2 - 2} \ dt \\ &=& {1 \over 2\sqrt 2}\ln\left({t + 1/t -\sqrt2 \over t + 1/t +\sqrt 2} \right) + C \end{eqnarray}$

Answer 4

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\overbrace{\color{#66f}{\large% \int{1 \over \root{x^{2} - 1}\pars{x^{2} + 1}}\,\dd x}} ^{\ds{\dsc{x} = \dsc{\cosh\pars{t}}}}\ =\ \int{1 \over \underbrace{\root{\cosh^{2}\pars{t} - 1}}_{\dsc{\sinh\pars{t}}}\ \bracks{\cosh^{2}\pars{t} + 1}}\,\sinh\pars{t}\,\dd t \\[5mm]&=\int{\dd t \over \cosh^{2}\pars{t} + 1} =\int{\sech^{2}\pars{t}\,\dd t \over 1 + \sech^{2}\pars{t}} =\int{\sech^{2}\pars{t}\,\dd t \over 2 - \tanh^{2}\pars{t}} \\[5mm]&={1 \over 2\root{2}} \int\bracks{{1 \over \root{2} - \tanh\pars{t}} +{1 \over \root{2} + \tanh\pars{t}}}\sech^{2}\pars{t}\,\dd t \\[5mm]&={\root{2} \over 4}\, \ln\pars{1 + \tanh\pars{t}/\root{2} \over 1 - \tanh\pars{t}/\root{2}} ={\root{2} \over 2}\,\,{\rm arctanh}\pars{\tanh\pars{t} \over \root{2}} \\[5mm]&={\root{2} \over 2} \,\,{\rm arctanh}\pars{\root{\cosh^{2}\pars{t} - 1} \over \root{2}\cosh\pars{t}} =\color{#66f}{\large{\root{2} \over 2} \,\,{\rm arctanh}\pars{\root{x^{2} - 1} \over \root{2}x}} + \mbox{a constant} \end{align}

Answer 5

Your answer is off by a minus sign, the error was made in the following step:

$$\color{red}{2xdx=\frac{-2u}{(u^2-1)^2}}$$

I conjecture that you had forgotten to write $\mathrm du$ as you have included the correct differential in the remaining steps. The correct relation should’ve been

$$2x\mathrm dx=\frac{2u\mathrm du}{(u^2-1)^2}$$

since the derivative of $\frac{-1}{x}$ is $\frac1{x^2}$.

Apart from this error, both the answers are equivalent. This can be seen by using the division and the power property of logarithms:

$$\log(x^n)=n\log x$$ $$\log\left(\frac{x}{y}\right)=\log x-\log y$$

As a result, $$\log x-\log y=\log(ax)-\log(ay)\forall a\in\mathbb R^+$$

So,

$$\frac{\log \left(\sqrt{2}+\frac{\sqrt{x^2-1}}{x}\right)-\log \left(\sqrt2-\frac{\sqrt{x^2-1}}{x}\right)}{2 \sqrt{2}}$$

$$=\frac{\log \left(x\sqrt{2}+\sqrt{x^2-1}\right)-\log \left(x\sqrt2-\sqrt{x^2-1}\right)}{2 \sqrt{2}}$$

$$=\frac{\log \left(x\sqrt{2}+\sqrt{x^2-1}\right)^2-\log \left(x\sqrt2-\sqrt{x^2-1}\right)^2}{2 \sqrt{2}}$$

$$=\frac{\log \left(3 x^2-1+2 \sqrt{2} \sqrt{x^2-1} x\right)-\log \left(3 x^2-1-2 \sqrt{2} \sqrt{x^2-1} x\right)}{4 \sqrt{2}}$$

$$=\frac{\log \left(-3 x^2-2 \sqrt{2} \sqrt{x^2-1} x+1\right)-\log \left(-3 x^2+2 \sqrt{2} \sqrt{x^2-1} x+1\right)}{4 \sqrt{2}}$$

Answer 6

Substituting $x=\frac1{t}$, the integral becomes:

$$\begin{align}\int\frac{\mathrm dx}{(x^2+1)\sqrt{x^2-1}}&=-\text{sgn}x\int\frac{t\mathrm dt}{(t^2+1)\sqrt{1-t^2}}\\&\overset{u=\sqrt{1-t^2}}{=}\text{sgn}x\int\frac{\mathrm du}{2-u^2}\\&=\frac{\text{sgn}x}{2\sqrt2}\ln\left|\frac{\sqrt2+u}{\sqrt2-u}\right|+C\\&=\frac{\text{sgn}x}{2\sqrt2}\ln\left|\frac{\sqrt2|x|+\sqrt{x^2-1}}{\sqrt2|x|-\sqrt{x^2-1}}\right|+C\\&=\frac1{\sqrt2}\tanh^{-1}\left(\frac{\sqrt{x^2-1}}{x\sqrt2}\right)+C\end{align}$$