MIT Integration Bee 2016 Number 20 $\int_0^{\infty}\frac{dx}{2+\cosh (x)}$ $$\int_0^{\infty}\frac{dx}{2+\cosh (x)}$$
I'm not sure how to approach this problem. I tried substituting $\cosh (x)  = \frac{e^x + e^{-x}}{2}$, and following the substitution $y = e^x$, I ended up with the integral 
$$2 \cdot \int_1^{\infty} \frac{1}{(y+2)^2-3} dy$$
Which I wasn't able to evaluate, nor am I sure it is the best form to proceed. 
Thanks!
 A: Using your substitution I get
$$2\int_1^\infty\frac{dy}{y^2+4y+1}=
2\int_3^\infty\frac{du}{u^2-3}.$$
This can be done by partial fractions.
A: An alternative approach for interest. With this integral it's relatively easy to rewrite $\cosh$ in terms of e, but in cases where it's less feasible, you can use the half-angle substitution for the hyperbolic tangent. 
Using that, we have, 
$$t = \tanh \frac x 2$$
$$\mathrm dx = \frac 2 {1-t^2} \mathrm dt$$
$$\cosh x = \frac{1+t^2}{1-t^2}$$
So for our limits we have, 
$$x = 0 \implies t = \tanh(0) = 0$$
$$x \to \infty \implies t = \lim_{t_0 \to \infty} \tanh\left(t_0\right) = 1$$
And for our integrand we have, 
$$\int_0^1 \frac {\mathrm dt} {2 + \frac {1+t^2}{1-t^2}} \cdot \frac 2 {1-t^2}$$
$$2\int_0^1 \frac {\mathrm dt} {2 - 2t^2 + 1 + t^2}$$
$$2\int_0^1 \frac {\mathrm dt} {3 - t^2}$$
Which is easily done with partial fractions, or another substitution. You will arrive at the same result as the others. If you want derivations of the above, I can edit this answer to include them. 
A: write your Integrand in the form
$$\frac{2e^x}{e^{2x}+4e^x+1}$$ and set $$e^x=t$$
A: $$\int_0^{\infty}\frac{dx}{2+\cosh (x)}=\int_0^{\infty}\frac{dx}{2+\frac{e^x+e^{-x}}{2}}$$
substitute $e^x=y\to x =\log y\to dx =\dfrac{dy}{y}$
limits become $x=0\to y=1;\;x\to\infty,\;y\to\infty$ so the integral becomes
$$\int_1^{\infty}\frac{dy}{2+\frac{y+\frac{1}{y}}{2}}\,\dfrac{1}{y}=\int_1^{\infty}\frac{2dy}{4y+y^2+1}=\int_1^{\infty}\frac{2dy}{4y+y^2+4-3}=\int_1^{\infty}\frac{2dy}{(y+2)^2-3}=$$
$$=2\int_1^{\infty}\frac{dy}{(y+2+\sqrt 3)(y+2-\sqrt 3)}=\dfrac{1}{\sqrt 3}\int_1^{\infty}\left(\frac{1}{y+\sqrt{3}+2}-\frac{1}{y-\sqrt{3}+2}\right)\,dy=$$
$$=\frac{1}{\sqrt 3}\left(\lim_{M\to\infty}\left[\log(M+\sqrt{3}+2)-\log(M-\sqrt{3}+2)\right]-\log\dfrac{1}{3+\sqrt 3}+\log\dfrac{1}{3-\sqrt 3}\right)=$$
$$=\frac{1}{\sqrt 3}\left(\lim_{M\to\infty} \log\dfrac{M+\sqrt{3}+2}{M-\sqrt{3}+2}+\log\frac{3+\sqrt 3}{3-\sqrt 3}\right)=\frac{\log(2+\sqrt 3)}{\sqrt 3}$$
Hope this helps
