$$ int _ { - infty } ^ { infty } \frac { e ^ { x } } { 1 + e ^ { 2 x } } d x $$

Question

Accepted Answer

# Explanation:
## Step1: Substitute $ t = e^{x} $
When $ t = e^{x} $, then $ dt=e^{x}dx $, and the integral $ \int_{-\infty}^{\infty}\frac{e^{x}}{1 + e^{2x}}dx=\int_{0}^{\infty}\frac{dt}{1 + t^{2}} $
## Step2: Use the integral formula $ \int\frac{1}{1 + t^{2}}dt=\arctan(t)+C $
$$
\begin{align*}
\int_{0}^{\infty}\frac{dt}{1 + t^{2}}&=\lim_{b
ightarrow\infty}\int_{0}^{b}\frac{dt}{1 + t^{2}}\
&=\lim_{b
ightarrow\infty}[\arctan(t)]_{0}^{b}\
&=\lim_{b
ightarrow\infty}(\arctan(b)-\arctan(0))
\end{align*}
$$
## Step3: Evaluate the limit
Since $ \lim_{b
ightarrow\infty}\arctan(b)=\frac{\pi}{2} $ and $ \arctan(0) = 0 $

# Answer:
$\frac{\pi}{2}$

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Question

Explanation:

Step1: Substitute \( t = e^{x} \)

Step2: Use the integral formula \( \int\frac{1}{1 + t^{2}}dt=\arctan(t)+C \)

Step3: Evaluate the limit

Answer: