# Solve integral $\int_0^\infty \frac{1}{\sqrt{x^2+1}(x^2+x+1)}dx$ using substitution $x=\sinh t$

How to solve the definite Integral $$\int_0^\infty \frac{1}{\sqrt{x^2+1}\ (x^2+x+1)}dx=\frac{\ln(2+\sqrt{3})}{\sqrt{3}}$$ with the hyperbolic function $$x=\sinh t$$?

Below is using the usual substitution $$x=\tan t$$ to do it. I thought it is the same because $$\sinh^2t+1=\cosh^2t$$ is similar with $$\tan^2t+1=\sec^2t$$. But, It isn't...

Substituting $$x=\tan t, dt=\sec^2{t} \ dx ,(0\leq t\leq\pi/2)$$ So, we get $$\int_0^\infty \frac{1}{\sqrt{x^2+1}(x^2+x+1)}dx =\int_{0}^{\frac{\pi}{2}}\frac{\sec{t}}{1+\tan{t}+\tan^2{t}}dt \\=\int_{0}^{\frac{\pi}{2}}\frac{\cos{t}}{1+\sin{t}\cos{t}}dt\\=I$$ Note that $$\cos{\left(\frac{\pi}{2}-t\right)}=\sin{t}$$, hence substituting $$t=\frac{\pi}{2}-\theta$$, we have $$\cos{\left(\frac{\pi}{2}-t\right)}=\sin{t}$$, hence substituting $$t=\frac{\pi}{2}-\theta$$, we have $$\int_{0}^{\frac{\pi}{2}}\frac{\cos{t}}{1+\sin{t}\cos{t}}dt=\int_{0}^{\frac{\pi}{2}}\frac{\sin{\theta}}{1+\sin{\theta}\cos{\theta}}d\theta=I$$ Therefore, by substitution $$\sin{t}-\cos{t}=\xi, (\sin{t}+\cos{t})dt=d\xi$$ $$2I=\int_{0}^{\frac{\pi}{2}}\frac{\cos{t}+\sin{t}}{1+\sin{t}\cos{t}}dt\\=\int_{-1}^{1}\frac{1}{1+\frac{1-\xi^2}{2}}d\xi \\=\int_{-1}^{1}\frac{1}{3-\xi^2}d\xi \\=\frac{1}{\sqrt3}\int_{-1}^{1}\frac{1}{\sqrt3-\xi}+\frac{1}{\sqrt3+\xi}d\xi \\=\frac{1}{ \sqrt3 }\left(\ln(\sqrt3-\xi)-\ln(\sqrt3+\xi) \bigg\rvert_{-1}^{1}\right)\\=\frac{2\ln(2+\sqrt{3})}{\sqrt3}$$ Because of the result, given integral is $$I=\int_0^\infty \frac{1}{\sqrt{x^2+1}(x^2+x+1)}dx=\frac{\ln(2+\sqrt{3})}{\sqrt{3}}$$

What is actually different with this using the substitution $$x=\sinh t$$ instead?

• Why don't you use the Euler substitution? Commented Jul 2, 2019 at 17:25
• @Dr.SonnhardGraubner I don't know Euler substitution alot...., I'll try using it too!! But I'm curious about using hyperbolic function because i'm stuck with it... Commented Jul 2, 2019 at 17:29

It is another route to reach the same result, as shown below \begin{align} &\int_0^\infty \frac{1}{\sqrt{x^2+1}\ (x^2+x+1)}\overset{x\to 1/x}{dx}\\ =&\ \frac12\int_0^\infty \frac{1+x}{\sqrt{x^2+1}\ (x^2+x+1)} \overset{x=\sinh t}{dx}\\ =& \ \frac12\int_0^\infty \frac{1+\sinh t}{\cosh^2 t+\sinh t}dt = \frac12 \int_0^\infty \frac{\text{sech}\,t(\text{sech}\,t +\tanh t)}{1+\text{sech}\,t\tanh t}dt\\ =& \int_0^\infty \frac{d(\tanh t- \text{sech}\,t )}{3-(\tanh t- \text{sech}\,t)^2}= \frac{\ln(2+\sqrt{3})}{\sqrt{3}} \end{align}

Letting $$x=\tan \theta$$ yields $$I=\int_0^{\infty} \frac{1}{\sqrt{x^2+1}\left(x^2+x+1\right)} d x = \int_0^{\frac{\pi}{2}} \frac{\cos \theta}{1+\sin \theta \cos \theta} d \theta$$ Multiplying both numerator and denominator by $$1-\sin \theta \cos \theta$$ yields $$I=\int_0^1 \frac{d s}{1-s^2\left(1-s^2\right)}-\int_0^1 \frac{c^2 d c}{1-\left(1-c^2\right) c^2}$$ Since $$s=\sin \theta$$ and $$c=\cos \theta$$ are dummy, we can rewrite $$I$$ as \begin{aligned} I & =\int_0^1 \frac{1-t^2}{t^4-t^2+1} d t \\ & =\int_0^1 \frac{\frac{1}{t^2}-1}{t^2+\frac{1}{t^2}-1} d t \\ & = -\int_0^1 \frac{d\left(t+\frac{1}{t}\right)}{\left(t+\frac{1}{t}\right)^2-3} \\ & = \frac{1}{2 \sqrt{3}} \ln \left|\frac{t+\frac{1}{t}+\sqrt{3}}{t+\frac{1}{t}-\sqrt{3}}\right|_0^1\\ & =\frac{1}{\sqrt{3}} \ln (2+\sqrt{3}) \end{aligned}

Hint: Substituting $$\sqrt{x^2+1}=x+t$$ we get $$x=\frac{1-t^2}{2t}$$ so $$dx=-\frac{t^2+1}{2 t^2}dt$$ and $$\sqrt{x^2+1}=\frac{1+t^2}{2t}$$

• This is about euler substitution? It's cool!! Commented Jul 2, 2019 at 17:40
• Yes, i have seen the question. Commented Jul 2, 2019 at 17:41
• so given integral changes $\int_{0}^{\infty}dx \to \int_{-1}^{-\infty}dt$??? Commented Jul 2, 2019 at 18:03
• Compute at first the indefinite integral Commented Jul 2, 2019 at 18:07
• But it isn't hard to compute indefinite integral? I get it but it is too coplicate to calculate.... Commented Jul 2, 2019 at 18:28

First simplify the integrand by substituting $$x=\dfrac{1-y}{1+y}$$:

\begin{align*} I &= \int_0^\infty \frac{dx}{\sqrt{x^2+1} \left(x^2+x+1\right)} \, dx \\ &= \sqrt2 \int_{-1}^1 \frac{1+y}{\sqrt{1+y^2} \left(3+y^2\right)} \, dy \\ &= \sqrt2 \int_{-1}^1 \frac{dy}{\sqrt{1+y^2} \left(3+y^2\right)} \end{align*}

Proceeding with the suggested substitution $$y=\sinh t$$, we have

$$\newcommand{\sech}{\operatorname{sech}} I = \sqrt2 \int_{-\sinh^{-1}1}^{\sinh^{-1}1} \frac{dt}{3+\sinh^2t} \cdot \frac{\sech^2t}{\sech^2t} = \sqrt2 \int_{-\sinh^{-1}1}^{\sinh^{-1}1} \frac{d\tanh t}{3-2\tanh^2t} \, dt$$

which can be further wrapped up with $$u=\tanh t$$,

$$I = \sqrt2 \int_{-\tfrac1{\sqrt2}}^{\tfrac1{\sqrt2}} \frac{du}{3-2u^2} = \boxed{\frac{\log\left(2+\sqrt3\right)}{\sqrt3}}$$