Find the integral $\int\frac{\sqrt{x-1}-\sqrt{x+1}}{\sqrt{x-1}-3\sqrt{x+1}}dx$

Question

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.

Answer 1

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

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

Answer 2

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 $e^{\pm t} = x \pm \sqrt{x^2-1}$

Answer 3

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.