# Evaluating an indefinite integral with a square root in the denominator [closed]

Hello it is simple but I can not evaluate the following integral:

$$\int \frac{1} {\sqrt{x^2-2x-8}}\,\mathrm{d}x$$

Give me a clue or show step by step solutions please.

Thank you very much.

## closed as off-topic by Did, Adrian Keister, Taroccoesbrocco, Xander Henderson, Lord Shark the UnknownAug 2 '18 at 5:26

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$$\int \left(\frac{1}{\sqrt{x^2-2x-8}}\right)dx=\int \left(\frac{1}{\sqrt{(x-1)^2-9}}\right)dx$$

Substitute: $u=x$ and $du=dx$:

$$\int \left(\frac{1}{\sqrt{(u)^2-9}}\right)du$$

Substitute: $u=3\sec(s)$ and $du=3\tan(s)\sec(s)ds$. Then $\sqrt{u^2-9}=\sqrt{9\sec^2(s)-9}=3\tan(s)$ and $s=\sec^{-1}\left(\frac{u}{3}\right)$

$$3\int \frac{sec(s)}{3}dx=\int \sec(s)dx=\ln(\tan(s)+\sec(s))+c=$$

$$\ln(\tan(\sec^{-1}\left(\frac{u}{3}\right))+\sec(\sec^{-1}\left(\frac{u}{3}\right)))+c=$$

$$\ln(\tan(\frac{x-1}{3})+\sec(\frac{x-1}{3}))+c=$$

$$\ln\left(-\sqrt{x^2-2x-8}-x+1\right)+c$$

• What's your justification for going from $\ln |stuff|$ to $\ln (stuff)$? – Zain Patel Jul 10 '15 at 13:12
• We don't have to! – Jan Jul 10 '15 at 13:14
• The argument of your logarithm is negative for quite some value of $x$. Or am I missing something, I most probably am? :-) – Zain Patel Jul 10 '15 at 13:15
• sure it is, but what you expected is not nesecarry – Jan Jul 10 '15 at 13:17

Hint: Complete the square and rewrite it as $\int \frac{1}{\sqrt{(x-1)^2-9}} dx$. Do a substitution $u=x-1$ and $du=dx$, then use trig substitution. I hope you can carry it on from there.

Use the substitution $u = x-1$, $\mathrm{d}u = \mathrm{d}x$ to get to $$\int \frac{\mathrm{d}u}{\sqrt{u^2-9}}$$ then apply the substitution $v = 3 \sec v \implies \mathrm{d}v = 3\sec v\tan v \, \mathrm{d}v$ to get to $$\int \sec v \, \mathrm{d}v$$ can you continue from there?

Rewrite in canonical form: $$I=\int\frac{\mathrm d\mkern1mu x}{\sqrt{x^2-2x-8}}=\int\frac{\mathrm d\mkern1mu x}{\sqrt{(x-1)^2-9}}=\int\frac{\mathrm d\mkern1mu x}{3\sqrt{\Bigl(\dfrac{x-1}3\Bigr)^2-1}}.$$ Now set $u=\dfrac{x-1}3$; you obtain: \begin{align*} I&=\int\frac{\mathrm d\mkern1mu u}{\sqrt{u^2-1}}=\operatorname{arcosh}u=\ln(u+\sqrt{u^2-1})\\ &=\operatorname{arcosh}\frac{x-1}3=\ln\Bigl(x-1+\sqrt{x^2-2x-8}\Bigr) \end{align*} modulo a constant.

Use $x \to \frac{4u^2+2}{u^2-1}$, ($u=\frac{\sqrt{x^2-2x-8}}{x-4}$) to have your integral (after simplification) in the following form $$2\int \frac{1}{1-u^2}du=\int \Big(\frac{1}{u+1}-\frac{1}{u-1}\Big)du$$

As the given $$\int \frac{1} {\sqrt{x^2-2x-8}}.dx$$ $$= \int \frac{1}{\sqrt{(x-1)^2-9}}$$ $$= \int \frac{1}{\sqrt{(x-1)^2-(3)^2}}$$ $$= \cosh^{-1}\frac{(x-1)}{3}$$