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My professor gave us an integral but I think it's incorrect. Is this correct?

$$ \DeclareMathOperator\arctanh{arctanh} \DeclareMathOperator\arccoth{arccoth} \int\frac{\mathrm d \, x}{x^2-1}= \arctanh ~x + C = \arccoth ~x + K$$

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    You can use [Wolfram Alpha](https://www.wolframalpha.com/) for such verifications. – 2017-01-05
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    Well first of all, there should be a $+C$ – 2017-01-05
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    and second we get $$-arctanh(x)+C$$ – 2017-01-05
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    @WiCK3DPOiSON i tried but i don't know how to do it :( – 2017-01-05
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    @Sarah type in `integral of dx/(x^2-1)` ... Wolfram Alpha is pretty forgiving. – 2017-01-05
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    You can check by taking the derivatives. – 2017-01-05

4 Answers 4

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First note that

$$\int\frac{1}{1-x^2}\ \mathrm dx= \operatorname{artanh}(x)+C$$

For the posted integral we have $$\int\frac{1}{x^2-1}\ \mathrm dx=-\int\frac{1}{1-x^2}\ \mathrm dx $$ $$=-\operatorname{artanh}(x)+C$$ So no, the stated equality is not correct. It's possible that either you or your professor, may have gotten mixed up.

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$\int \frac{1}{1-x^2} \,dx = \arctanh x + c$

Also,

$\int \frac{1}{1-x^2} \,dx = \arccoth x + c$

If you are writing as written above then change 1 into -1.

Or $\int \frac{1}{x^2-1} \,dx = -\int \frac{1}{1-x^2} \,dx -\arctanh x + c$

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    @Sarah its not good. You accepted my answer and then unaccepted it. My answer and the new accepted answer has same explanation. – 2017-01-05
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It is useful to rewrite $arccoth x$ as an inverse function. In this case, let $f(x)=cothx=\frac{e^x+e^{-x}}{e^x-e^{-x}}=\frac{e^{2x}+1}{e^{2x}-1}$. To determine the inverse, just solve for $x$ in the equation $y=\frac{e^{2x}+1}{e^{2x}-1}$.

\begin{align*} y&=\frac{e^{2x}+1}{e^{2x}-1}\\ e^{2x}y-y&=e^{2x}+1\\ e^{2x}(y-1)&=y+1\\ e^{2x}&=\frac{y+1}{y-1}\\ 2x&=\ln\left(\frac{y+1}{y-1}\right)\\ x&=\frac{1}{2}\ln\left(\frac{y+1}{y-1}\right)\\ arccothx&=\frac{1}{2}\ln\left(\frac{x+1}{x-1}\right) \end{align*} You will get something similar for $arctanhx$! \begin{equation*} arctanhx=\frac{1}{2}\ln\left(\frac{x+1}{1-x}\right) \end{equation*} For the integral, use partial fractions. The rewrite is \begin{equation*} \frac{1}{x^2-1}=\frac{1}{2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right) \end{equation*}

\begin{equation*} \int\frac{1}{x^2-1}\mathrm{d}x=\frac{1}{2}\left(\ln\left|\frac{x-1}{x+1}\right|\right)+C \end{equation*}

Your lecturer was close! for $|x|>1$, we have $\int\frac{1}{x^2-1}\mathrm{d}x=-arccothx+C$ and for $|x|<1$, we have $\int\frac{1}{x^2-1}\mathrm{d}x=-arctanhx+C$.

I hope this helps (and is all correct)!

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By definition,

$$\text{arcoth }x=\text{artanh }\frac1x.$$

Then taking the derivative,

$$\text{arcoth }'x=-\frac1{x^2}\text{artanh }'\frac1x==-\frac1{x^2}\frac1{1-\dfrac1{x^2}}=\frac1{x^2-1}.$$

so there is a sign error.