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

Below is using $\tan x$ to do it. I thought it is same with this because $\sinh^2(x)+1=\cosh^2(x)$ is similar with $\tan^2(x)+1=\sec^2(x)$ ,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
$$I=\int_0^\infty \frac{1}{\sqrt{x^2+1}(x^2+x+1)}dx=\frac{\ln(2+\sqrt{3})}{\sqrt{3}}$$

And,what is actually different with this and using $\sinh(x)$?

\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}
