Integral $I=\int \frac{dx}{(x^2+1)\sqrt{x^2-4}} $

Question

Frankly, i don't have a solution to this, not even incorrect one, but, this integral looks a lot like that standard type of integral $I=\int\frac{Mx+N}{(x-\alpha)^n\sqrt{ax^2+bx+c}}$ which can be solved using substitution $x-\alpha=\frac{1}{t}$ so i tried to find such subtitution that will make this integral completely the same as this standard integral so i could use substitution i mentioned, so i tried two following substitutions

$x^2-4=t^2 \Rightarrow x^2=t^2+4 \Rightarrow x=\sqrt{t^2+4}$ then i had to determine $dx$

$2xdx=2tdt \Rightarrow dx=\frac{tdt}{\sqrt{t^2+4}}$

from here i got:

$\int\frac{dt}{(t^2+5)\sqrt{(t^2+4)}}$ but i have no idea what could i do with this, so i tried different substitution

$x^2+1=t^2$ and then, by implementing the same pattern i used with the previous substitution i got this integral

$\int\frac{dt}{t\sqrt{(t^2-1)(t^2-5)}}$ but again, i don't know what to do with this, so i could use some help.

Answer 1

Use $x=2\cosh t$, so the integral becomes $$ \int\frac{1}{1+4\cosh^2t}\,dt= \int\frac{1}{1+e^{2t}+2+e^{-2t}}= \int\frac{e^{2t}}{e^{4t}+3e^{2t}+1}\,dt $$ and then set $u=e^{2t}$ so you get $$ \frac{1}{2}\int\frac{1}{u^2+3u+1}\,du $$ that can be computed by partial fractions.

Answer 2

HINT:

Your integral:

$$\text{I}=\int\frac{1}{\left(x^2+1\right)\sqrt{x^2-4}}\space\text{d}x=$$

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Subsitute $x=2\sec(u)$ and $\text{d}x=2\tan(u)\sec(u)\space\text{d}u$.

Then $\sqrt{x^2-4}=\sqrt{4\sec^2(u)-4}=2\tan(u)$ and $u=\text{arcsec}\left(\frac{x}{2}\right)$:

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$$\int\frac{\sec(u)}{4\sec^2(u)+1}\space\text{d}u=\int\frac{\cos(u)}{5-\sin^2(u)}\space\text{d}u=$$

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Substitute $s=\sin(u)$ and $\text{d}s=\cos(u)\space\text{d}u$:

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$$\int\frac{1}{5-s^2}\space\text{d}s=\frac{1}{5}\int\frac{1}{1-\frac{s^2}{5}}\space\text{d}s=$$

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Substitute $p=\frac{s}{\sqrt{5}}$ and $\text{d}p=\frac{1}{\sqrt{5}}\space\text{d}s$:

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$$\frac{1}{\sqrt{5}}\int\frac{1}{1-p^2}\space\text{d}p=\frac{\text{arctanh}\left(p\right)}{\sqrt{5}}+\text{C}=$$ $$\frac{\text{arctanh}\left(\frac{s}{\sqrt{5}}\right)}{\sqrt{5}}+\text{C}=\frac{\text{arctanh}\left(\frac{\sin\left(u\right)}{\sqrt{5}}\right)}{\sqrt{5}}+\text{C}=$$ $$\frac{\text{arctanh}\left(\frac{\sin\left(\text{arcsec}\left(\frac{x}{2}\right)\right)}{\sqrt{5}}\right)}{\sqrt{5}}+\text{C}=\frac{\text{arctanh}\left(\frac{\sqrt{x^2-4}}{x\sqrt{5}}\right)}{\sqrt{5}}+\text{C}$$

Answer 3

The curve $t^2=x^2-4$ is a hyperbola, which can be parametrized by a single value. Rewrite this equation as $(x+t)(x-t)=4$, and set $y=x+t$. Then $4/y=x-t$, and we have $$ x=\frac{y+\frac{4}{y}}{2},\;\;\;\;t=\frac{y-\frac{4}{y}}{2}. $$ Compute $dx=(1/2-2 y^{-2})dy$, and the integral becomes $$ \int \frac{dx}{(x^2+1)\sqrt{x^2-4}}=\int \frac{\frac{1}{2}-\frac{2}{y^2}}{\left(\left(\frac{y+\frac{4}{y}}{2}\right)^2+1\right)\left(\frac{y-\frac{4}{y}}{2}\right)}dy=\int \frac{4y\, dy}{y^4+12 y^2+16}. $$ This can be computed using partial fractions.

Answer 4

An approach that does not involve trigonometric substitutions or partial fractions is to substitute $x=\frac1{t}$:

$$\mathcal I=\int\frac{\mathrm dx}{(x^2+1)\sqrt{x^2-4}}$$ $$=\text{sgn}x\int\frac{t\mathrm dt}{(t^2+1)\sqrt{1-4t^2}}$$

Now, substitute $\sqrt{1-4t^2}=u\implies-8t\mathrm dt=2u\mathrm du$:

$$\mathcal I=\frac{-\text{sgn}x}4\int\frac{\mathrm du}{1+\left(\frac{1-u^2}{4}\right)}$$

You can take it from here.

Answer 5

I’m surprised that nobody thought of substituting $x=\tan t$:

$$\mathcal I=\int\frac{\mathrm dx}{(x^2+1)\sqrt{x^2-4}} $$ $$=\int\frac{\mathrm dt}{\sqrt{\tan^2t-4}}$$ $$=\int\frac{\cos t\mathrm dt}{\sqrt{5\sin^2t-4}}$$

It is obvious what to do next.

Answer 6

In general, let $t=\frac x{\sqrt{x^2+b}}$

$$\int \frac{1}{(x^2+a)\sqrt{x^2+b}} dx = \int \frac1{(b-a)t^2+a} dt $$ and, in particular $$\int \frac{1}{(x^2+1)\sqrt{x^2-4}} dx = \int \frac1{1-5t^2} dt=\frac1{\sqrt5}\coth^{-1}\sqrt5 t $$