# Computation of the definite integral $\int _{\log _e\left(8\right)}^{\infty }\sqrt{\frac{e^x+1}{e^x-3}}\:$

Good evening to everyone! I have the following integral: $$\int _{\log _e\left(8\right)}^{\infty }\sqrt{\frac{e^x+1}{e^x-3}}\:$$ and I don't know how to study its convergence. Here's what I tried: $$\int _{\log _e\left(8\right)}^{\infty }\sqrt{\frac{e^x+1}{e^x-3}}\,dx\: = \int _{8}^{\infty }\frac {1}{u}\sqrt{\frac{u+1}{u-3}}du\:$$ And I don't know what to do from here. Thanks for any response.

• @amcalde thanks for telling me I just corrected it Jun 30 '16 at 17:24
• You can use partial fractions. But I think there is a mistake it stud be $U^3-3u^2$ in the denominator Jun 30 '16 at 17:30
• Maybe the convergence of the integral is something to consider in first; in particular, I do not see why the change of variables is so important here. Jun 30 '16 at 17:38

It's not necessary to change variables. Note that $e^x+1\gt e^x-3$, so
$$\sqrt{e^x+1\over e^x-3}\gt1\quad\text{for }x\gt\ln3$$
$$\int_{\ln8}^\infty\sqrt{e^x+1\over e^x-3}dx\gt\int_{\ln8}^\infty dx$$
Let $e^x-3=u$, hence $$e^x\text{d}x=\text{d}u\implies \text{d}x=\frac{\text{d}u}{u+3} \quad \textrm{and} \quad e^x+1=u+4$$ thus your integral becomes $$\int\limits_{5}^{\infty}\frac{1}{u+3}\sqrt{\frac{u+4}{u}}\text{d}u$$Now, you can write that $\displaystyle 1\le \sqrt{\frac{u+4}{u}}=\sqrt{1+\frac{4}{u}}$, so $$\int\limits_{5}^{\infty}\frac{1}{u+3}\text{d}u\le\int\limits_{5}^{\infty}\frac{1}{u+3}\sqrt{1+\frac{4}{u}}\text{d}u$$ so, the integral does not converge.