# Solve an integral $\int\frac{1}{(1+x^2)\sqrt{x^2-1}}dx$

Solve an integral $$\int\frac{1}{(1+x^2)\sqrt{x^2-1}}dx$$

Using the third substitution of Euler, $$\sqrt{x^2-1}=(x-1)t\Rightarrow x=\frac{1+t^2}{1-t^2},dx=\frac{4t}{(1-t^2)^2}$$

we get an integral $$\int\frac{1-t^2}{1+t^4}dt=\frac{1}{2}\int\frac{1-\sqrt 2 t}{t^2-\sqrt 2 t+1}dt+\frac{1}{2}\int\frac{1+\sqrt 2t}{t^2+\sqrt 2t+1}dt$$

Using substitutions $$u=t^2-\sqrt 2t+1$$ and $$v=t^2+\sqrt 2t+1$$ we get $$\int\frac{1-t^2}{1+t^4}dt=-\frac{\sqrt 2}{4}\ln|u|+\frac{\sqrt 2}{4}\ln|v|=-\frac{\sqrt 2}{4}\ln|t^2-\sqrt 2t+1|+\frac{\sqrt 2}{4}\ln|t^2+\sqrt 2t+1|$$

From $$x=\frac{1+t^2}{1-t^2}\Rightarrow t=\sqrt{\frac{x-1}{x+1}}\Rightarrow$$

$$\int\frac{1}{(1+x^2)\sqrt{x^2-1}}dx=$$$$-\frac{\sqrt 2}{4}\ln\left|\frac{x-1}{x+1}-\sqrt 2\sqrt{\frac{x-1}{x+1}}+1\right|+\frac{\sqrt 2}{4}\ln\left|\frac{x-1}{x+1}+\sqrt 2\sqrt{\frac{x-1}{x+1}}+1\right|+c$$

Is there another, quicker method to solve this integral, rather than Euler substitution.

• You could use hyperbolic functions and substitute in $x=\cosh t$. Then use the substitution $u=\tanh t$, which yields the solution. – S.C.B. Mar 5 '17 at 9:24
• @S.C.B. How did you know that substitution would work? Did you actually work it through, or just intuition/recognition? – mrnovice Mar 5 '17 at 9:28
• @mrnovice Are the two necessarily distinct? It's both. – S.C.B. Mar 5 '17 at 9:29
• @S.C.B. Well the former requires more trial and error, whilst the latter can be done by inspection – mrnovice Mar 5 '17 at 9:30
• Try $\frac 1 {\sin t} = x$ substitution, it's working. – user261263 Mar 5 '17 at 9:33

Here's what I tried. Setting $x=\sec \theta$, we get

\begin{align*} \int \frac{1}{(x^2+1)\sqrt{x^2-1}} \mathrm{d}x &= \int \frac{\sec \theta \tan \theta}{(\sec^2 \theta+1)\sqrt{\sec^2 \theta-1}} \mathrm{d}\theta \\ &= \int \frac{\sec \theta}{\sec^2 \theta+1} \mathrm{d}\theta \\ &= \int \frac{\cos \theta}{\cos^2 \theta+1} \mathrm{d}\theta \\ &= \int \frac{1}{2-\sin^2 \theta} \mathrm{d}(\sin \theta)\\ &= \frac{1}{2\sqrt 2} \ln \left|\frac{\sqrt 2 + \sin \theta}{\sqrt 2 - \sin \theta}\right|+C \end{align*}

Since, $\sec \theta = x \implies \sin \theta = \frac{\sqrt {x^2-1}}{x}$, we get

$$\int \frac{1}{(x^2+1)\sqrt{x^2-1}} \mathrm{d}x = \frac{1}{2\sqrt 2} \ln \left|\frac{\sqrt 2x + \sqrt{x^2-1}}{\sqrt 2x - \sqrt{x^2-1}}\right| +C$$

Setting $$x=\sinh(t)$$ then we have $$dx=\cosh(t)dt$$ and our integral will be $$\int \frac{dt}{1+\sinh^2(t)}$$ then use $$\sinh(t)=-2\,{\frac {\tanh \left( t/2 \right) }{ \left( \tanh \left( t/2 \right) \right) ^{2}-1}}$$ note that $$x^2-1=\sinh^2(t)-1=\cosh^2(t)$$

• How do you simplify $\;\sqrt{x^2-1}$, initially? – Bernard Mar 5 '17 at 10:30

\begin{align}\int \frac{1}{(1+x^2)\sqrt{x^2 - 1}} \, dx &= \int \dfrac{\sinh u}{(\cosh^2 u + 1)\sqrt{\cosh^2 u - 1}} \, du && x = \cosh u \\ &= \int \dfrac{\sinh u}{(\cosh^2 u + 1)\sinh u} \, du \\ &= \int \dfrac{1}{\cosh^2 u + 1}\, du \\ && \end{align} After which things become simple in terms of hyperbolic tsngent of u and it's inverse. Unsubstitute after that.

$I=\int\frac{1}{(1+x^2)\sqrt{x^2-1}}dx$
Put $x=\frac{1}{t}\implies dx=-\frac{1}{t^2}dt \implies I=-\int\frac{t}{(t^2+1)\sqrt{1-t^2}}dt$ Put$1-t^2=y^2\implies -t dt=y dy. \implies I=\int\frac{y}{(2-y^2)y}dy =\frac{1}{2\sqrt2}ln\frac{\sqrt2+y}{\sqrt2-y}+C =\frac{1}{2\sqrt2}ln\frac{\sqrt2+\sqrt{1-t^2}}{\sqrt2-\sqrt{1-t^2}}+C =\frac{1}{2\sqrt2}ln\frac{\sqrt2x+\sqrt{x^2-1}}{\sqrt2x-\sqrt{x^2-1}}+C$