11. $$ \int_{0}^{2} \frac{e^{x}}{1+e^{2 x}} d x $$

Question

UpStudy AI · Accepted Answer

To evaluate the integral:

$$
I = \int_{0}^{2} \frac{e^{x}}{1 + e^{2x}} \, dx
$$

**Step 1: Substitution**

Let $$ u = e^x $$. Then, $$ du = e^x dx $$, or equivalently, $$ dx = \frac{du}{u} $$.

**Step 2: Change of Limits**

When $$ x = 0 $$:
$$
u = e^0 = 1
$$

When $$ x = 2 $$:
$$
u = e^2
$$

**Step 3: Rewrite the Integral**

Substituting $$ u $$ and $$ dx $$ into the integral:

$$
I = \int_{1}^{e^2} \frac{u}{1 + u^2} \cdot \frac{du}{u} = \int_{1}^{e^2} \frac{1}{1 + u^2} \, du
$$

**Step 4: Integrate**

The integral of $$ \frac{1}{1 + u^2} $$ is $$ \arctan(u) $$:

$$
I = \left[ \arctan(u) \right]_{1}^{e^2} = \arctan(e^2) - \arctan(1)
$$

Since $$ \arctan(1) = \frac{\pi}{4} $$:

$$
I = \arctan(e^2) - \frac{\pi}{4}
$$

**Final Answer:**

$$
\boxed{\,\arctan\left(e^{2}\right)\;-\;\dfrac{\pi}{4}\,}
$$
The integral evaluates to $$ \arctan(e^{2}) - \frac{\pi}{4} $$.

11. \( \int_{0}^{2} \frac{e^{x}}{1+e^{2 x}} d x \)