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

Question

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?

Answer 1

Substitute $x=\sinh t $ as follows \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}

Answer 2

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

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

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

$$\newcommand{\sech}{\operatorname{sech}} I = \sqrt2 \int_{-\sinh^{-1}1}^{\sinh^{-1}1} \frac{dt}{3+\sinh^2t} \cdot \frac{\sech^2t}{\sech^2t} = 2\sqrt2 \int_0^{\sinh^{-1}1} \frac{d(\tanh t)}{3-2\tanh^2t}$$

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

$$I = 2\sqrt2 \int_0^{\tfrac1{\sqrt2}} \frac{du}{3-2u^2} = \boxed{\frac{\log\left(2+\sqrt3\right)}{\sqrt3}}$$

Answer 3

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

Answer 4

As the approach of performing the substitution $x=\sinh t$ has been described by @Quanto beautifully, I will share an alternative way of solving the given definite integral. For this, we need to observe that we can obtain the derivative of $\frac1{\sqrt x}$ in the integrand.

Now, after performing the imminent substitution $t=\frac1{\sqrt x}$, the integral can be simplified by exploiting the symmetry of the integrand under the substitution $t\to\frac1t$.

$$\begin{align}\int_0^\infty\frac{\mathrm dx}{(x^2+x+1)\sqrt{x^2+1}}&=\int_0^\infty\frac{\mathrm dx}{x^\frac32\left(x+\frac1x+1\right)\sqrt{x+\frac1x}}\\&\overset{t=\frac1{\sqrt x}}{=}\int_0^\infty\frac{2\mathrm dt}{\left(t^2+\frac1{t^2}+1\right)\sqrt{t^2+\frac1{t^2}}}\\&\overset{t\to\frac1t}{=}\int_0^\infty\frac{\mathrm d\left(t-\frac1t\right)}{\left(\left(t-\frac1t\right)^2+3\right)\sqrt{\left(t-\frac1t\right)^2+2}}\\&\overset{t-\frac1t=\frac1y}{=}\frac1{3\sqrt2}\int_{-\infty}^\infty\frac{|y|\,\mathrm dy}{\left(y^2+\frac13\right) \sqrt{y^2+\frac12}}\\&=\frac{\sqrt2}3\int_0^\infty\frac{y\,\mathrm dy}{\left(y^2+\frac13\right) \sqrt{y^2+\frac12}}\\&\overset{v=\sqrt{y^2+\frac12}}{=}\frac{\sqrt2}3\int_{\frac1{\sqrt2}}^\infty\frac{\mathrm dv}{v^2-\frac16}\\&=\frac2{\sqrt3}\coth^{-1}\sqrt6v\bigg|_{\infty}^{\frac1{\sqrt2}}\\&=\frac2{\sqrt3}\coth^{-1}\sqrt3\\&=\boxed{\frac{\ln(2+\sqrt3)}{\sqrt3}}\end{align}$$

Answer 5

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