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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?

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  • $\begingroup$ Why don't you use the Euler substitution? $\endgroup$ Commented Jul 2, 2019 at 17:25
  • $\begingroup$ @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... $\endgroup$
    – user366725
    Commented Jul 2, 2019 at 17:29

4 Answers 4

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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}

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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 $$ \begin{aligned} I & =\int_0^1 \frac{1-t^2}{t^4-t^2+1} d t \\ & =\int_0^1 \frac{\frac{1}{t^2}-1}{t^2+\frac{1}{t^2}-1} d t \\ & = -\int_0^1 \frac{d\left(t+\frac{1}{t}\right)}{\left(t+\frac{1}{t}\right)^2-3} \\ & = \frac{1}{2 \sqrt{3}} \ln \left|\frac{t+\frac{1}{t}+\sqrt{3}}{t+\frac{1}{t}-\sqrt{3}}\right|_0^1\\ & =\frac{1}{\sqrt{3}} \ln (2+\sqrt{3}) \end{aligned} $$

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

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  • $\begingroup$ This is about euler substitution? It's cool!! $\endgroup$
    – user366725
    Commented Jul 2, 2019 at 17:40
  • $\begingroup$ Yes, i have seen the question. $\endgroup$ Commented Jul 2, 2019 at 17:41
  • $\begingroup$ so given integral changes $\int_{0}^{\infty}dx \to \int_{-1}^{-\infty}dt$??? $\endgroup$
    – user366725
    Commented Jul 2, 2019 at 18:03
  • $\begingroup$ Compute at first the indefinite integral $\endgroup$ Commented Jul 2, 2019 at 18:07
  • $\begingroup$ But it isn't hard to compute indefinite integral? I get it but it is too coplicate to calculate.... $\endgroup$
    – user366725
    Commented Jul 2, 2019 at 18:28
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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 $u=\tanh t$,

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

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