22. $\\int_{0}^{\\infty}\\frac{e^{x}}{e^{2x}+3}dx$

Explanation:

Step1: Substitution

Let (u = e^{x}), then (du=e^{x}dx). When (x = 0), (u = 1); as (x\rightarrow\infty), (u\rightarrow\infty). The integral becomes (\int_{1}^{\infty}\frac{du}{u^{2}+3}).

Step2: Use integral formula

Recall (\int\frac{du}{u^{2}+a^{2}}=\frac{1}{a}\arctan(\frac{u}{a})+C) ((a=\sqrt{3}) here). So (\int_{1}^{\infty}\frac{du}{u^{2}+3}=\lim_{b\rightarrow\infty}\int_{1}^{b}\frac{du}{u^{2}+3}). [ \begin{align*} \lim_{b\rightarrow\infty}\int_{1}^{b}\frac{du}{u^{2}+3}&=\lim_{b\rightarrow\infty}\left[\frac{1}{\sqrt{3}}\arctan\left(\frac{u}{\sqrt{3}}\right)\right]{1}^{b}\ &=\lim{b\rightarrow\infty}\left(\frac{1}{\sqrt{3}}\arctan\left(\frac{b}{\sqrt{3}}\right)-\frac{1}{\sqrt{3}}\arctan\left(\frac{1}{\sqrt{3}}\right)\right) \end{align*} ]

Step3: Evaluate limit

Since (\lim_{t\rightarrow\infty}\arctan(t)=\frac{\pi}{2}) and (\arctan\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}). [ \begin{align*} &\lim_{b\rightarrow\infty}\left(\frac{1}{\sqrt{3}}\arctan\left(\frac{b}{\sqrt{3}}\right)-\frac{1}{\sqrt{3}}\arctan\left(\frac{1}{\sqrt{3}}\right)\right)\ =&\frac{1}{\sqrt{3}}\left(\frac{\pi}{2}-\frac{\pi}{6}\right)\ =&\frac{1}{\sqrt{3}}\times\frac{\pi}{3}\ =&\frac{\pi}{3\sqrt{3}}=\frac{\sqrt{3}\pi}{9} \end{align*} ]

Answer:

(\frac{\sqrt{3}\pi}{9})