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Find the integral $$I=\int\dfrac{\sqrt{x-1}-\sqrt{x+1}}{\sqrt{x-1}-3\sqrt{x+1}}dx$$

My try: $$I=\int\dfrac{x-1-x-1}{\left(\sqrt{x-1}-3\sqrt{x+1}\right)\left(\sqrt{x-1}+\sqrt{x+1}\right)}dx\\=-2\int\dfrac{1}{x-1+\sqrt{x^2-1}-3\sqrt{x^2-1}-3x-3}dx\\=-2\int\dfrac{1}{-2x-4-2\sqrt{x^2-1}}dx\\=\int\dfrac{1}{x+\sqrt{x^2-1}+2}dx$$ I am not sure what to do now. I thought about integrating by parts but the first derivative of $\dfrac{1}{x+\sqrt{x^2-1}+2}$ is $\dfrac{-\sqrt{x^2-1}-x}{(x+\sqrt{x^2-1}+2)^2\sqrt{x^2-1}}$ which doesn't seem very clear.

SAQ
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  • try $\sqrt{x^2-1}=x-t$ – Math Student Jan 19 '24 at 20:24
  • @MathStudent That's not correct. $x^2 - 1 = (x-i)(x+i)$, not $(x-i)^2$. – Robert Israel Jan 19 '24 at 20:26
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    Try rationalizing the denominator, not the numerator: multiply top and bottom by $\sqrt{x-1} + 3 \sqrt{x+1}$ – Robert Israel Jan 19 '24 at 20:27
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    @RobertIsrael, what?! The first Euler substitution is applicable as $a=1>0$. Actually it would be better if we put $\sqrt{x^2-1}=t-x$. Then $x$ would be $x=\dfrac{t^2+1}{2t}$ in terms of $t$ and we know how to integrate rational functions... I suppose we're working in $\mathbb{R}$. – Math Student Jan 19 '24 at 20:29
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    meanwhile, not the quickest: multiply by $\frac{\sqrt{x+1}}{\sqrt{x+1}}$ and then take, say $x = \cosh t$ Then a half angle substitution (modified for hyperbolic functions) leads to partial fractions. – Will Jagy Jan 19 '24 at 20:35
  • @RobertIsrael, how is that different? Then I would need to know how to find $\int\dfrac{x}{4x+5}dx$ and $\int\dfrac{\sqrt{x^2-1}}{4x+5}dx$, which doesn't seem to be the case. Am I missing something? – SAQ Jan 19 '24 at 20:50

3 Answers3

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We can avoid trigonometric and Euler substitutions. The integrand is defined for $x\ge1$, and dividing through by $\sqrt{x+1}$ and substituting $y=\sqrt{\dfrac{x-1}{x+1}}$ (note that $y\ge0$) gives

$$\int \frac{\sqrt{\frac{x-1}{x+1}} - 1}{\sqrt{\frac{x-1}{x+1}} - 3} \, dx = \int \frac{4y}{(y-1)(y-3)(y+1)^2} \, dy$$

user170231
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    Nice solution! I would argue though that the Euler substitution here gives an easier rational function to integrate. Namely $\int\dfrac{t^2-1}{(t+2)t^2} dt, t=\sqrt{x^2-1}+x$ if I haven't messed up the calculations. Nice observation though! – Math Student Jan 19 '24 at 21:04
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From the step you left off, use $x = \cosh t$:

$$I = \int\frac{\sinh t }{\cosh t + \sinh t + 2}dt = \int\frac{\sinh t}{e^t+2}dt$$

$$= \frac{1}{8}\int 1-2e^{-t}+\frac{3e^t}{e^t+2}dt = \frac{1}{8}\left(t+2e^{-t}+3\log(e^t+2)\right)+C$$

$$= \frac{1}{8}\left(\cosh^{-1}x+2x-2\sqrt{x^2-1}+3\log\left(x+\sqrt{x^2-1}+2\right)\right)+C$$

by using polynomial long division, then $e^{\pm t} = x \pm \sqrt{x^2-1}$

Ninad Munshi
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Since $\frac{\sqrt{x-1}}{\sqrt{x+1}}=\tan{\left(\frac12\sec^{-1}(x)\right)}$, we use the transformation

$$\int f\left(\tan{\left(\frac12\csc^{-1}(x)\right)}, \tan{\left(\frac12\sec^{-1}(x)\right)} \right)\,dx=\int f\left(e^{i\cos^{-1}(x)}, \frac{1-e^{i\cos^{-1}(x)}}{1+e^{i\cos^{-1}(x)}}\right)\,dx,\tag{1}$$

valid for $x\geq1$. This transformation has been discussed here, here, and here.

We first divide both the numerator and denominator by $\sqrt{x+1}$:

$$ \int\dfrac{\sqrt{x-1}-\sqrt{x+1}}{\sqrt{x-1}-3\sqrt{x+1}}\,dx = \int\dfrac{\frac{\sqrt{x-1}}{\sqrt{x+1}}-1}{\frac{\sqrt{x-1}}{\sqrt{x+1}}-3}\,dx. $$

Now, we'll substitute $\frac{\sqrt{x-1}}{\sqrt{x+1}}$ with $\frac{1-e^{i\cos^{-1}(x)}}{1+e^{i\cos^{-1}(x)}}$ using the transformation $(1)$:

$$ = \int\dfrac{\frac{1-e^{i\cos^{-1}(x)}}{1+e^{i\cos^{-1}(x)}}-1}{\frac{1-e^{i\cos^{-1}(x)}}{1+e^{i\cos^{-1}(x)}}-3}\,dx. $$

Let's denote $u = \cos^{-1}(x)$. Then, $x = \cos{u}$ and $dx = -\sqrt{1-x^2}\,du = -\sqrt{1-\cos^2{(u)}}\, du = -\sin{(u)}\,du$. Now, we'll perform the substitution:

$$ = -\int\dfrac{\frac{1-e^{iu}}{1+e^{iu}}-1}{\frac{1-e^{iu}}{1+e^{iu}}-3}\sin{(u)}\,du. $$

Since $\sin(u) = \frac{i}{2}(e^{-iu} - e^{iu})$ and simplifying, we get:

$$ \frac{i}{2}\int \frac{e^{2iu}-1}{2e^{iu}+1} \, du. $$

Now, let $v = e^{iu}$. Then, $dv = ie^{iu}\,du = iv\,du$ and $du = \frac{1}{iv}\,dv$. Substituting, we have:

$$\frac12\int \frac{v^2 - 1}{v(2v + 1)} \, dv=\frac12\int 1\,dv - \frac12\int \frac{v+2}{v(2v+1)}\,dv.$$

At this point, we can perform partial fraction decomposition on the second integral on the right, which is even simpler than the integral @MathStudent arrived at through Euler substitution.

Emmanuel José García
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