Find the area bounded by the lines: y = (e^x)/(1 + e^2x)

Find the area bounded by the lines: $y = \frac{e^x}{1 + e^{2x}}$, $y = 0$, $x = 0$, $x = 1$.

We will substitute $u=e^x$ to reduce the integral to a standard arctangent form.

Step 1: Set up the integral

\[A = \int_0^1 \frac{e^x}{1 + e^{2x}} \, dx\]

Step 2: Use substitution

Let $u = e^x$, so $du = e^x \, dx$, giving $dx = \frac{du}{u}$. When $x=0$: $u=1$. When $x=1$: $u=e$. Also $e^{2x}=u^2$.

\[A = \int_1^e \frac{u}{1 + u^2} \cdot \frac{du}{u} = \int_1^e \frac{1}{1 + u^2} \, du\]

Step 3: Evaluate the integral

\[A = [\arctan u]_1^e = \arctan e - \arctan 1 = \arctan e - \frac{\pi}{4}\]

\[\boxed{A = \arctan e - \frac{\pi}{4} \text{ square units}}\]

Step 4: Numerical value

\[\arctan e \approx 1.2182 \text{ rad}, \qquad \frac{\pi}{4} \approx 0.7854 \text{ rad}\]

\[A \approx 1.2182 - 0.7854 \approx 0.433 \text{ square units}\]

Summary

\[\boxed{A = \arctan e - \frac{\pi}{4} \approx 0.433 \text{ square units}}\]

Key insight: The substitution $u=e^x$ transforms the exponential expression into the standard arctangent integral, so the result involves $\arctan e$ — a transcendental combination of the exponential and inverse trigonometric functions.