Evaluate the definite integral.

Evaluate the definite integral. 1 ex 1 + e2x dx 0

The Correct Answer and Explanation is:

We are tasked with evaluating the definite integral:∫01ex1+e2x dx\int_0^1 \frac{e^x}{1 + e^{2x}} \, dx∫01​1+e2xex​dx

To solve this, we first observe the structure of the integrand. Let’s proceed step-by-step:

Step 1: Simplifying the integrand

The given expression is:ex1+e2x\frac{e^x}{1 + e^{2x}}1+e2xex​

We can make a substitution to simplify the expression. Let:u=exu = e^xu=ex

Then, the differential of uuu is:du=ex dxdu = e^x \, dxdu=exdx

When x=0x = 0x=0, u=e0=1u = e^0 = 1u=e0=1, and when x=1x = 1x=1, u=e1=eu = e^1 = eu=e1=e.

Thus, the limits of integration change from x=0x = 0x=0 to x=1x = 1x=1, becoming u=1u = 1u=1 to u=eu = eu=e.

Step 2: Substitute into the integral

Now, substitute u=exu = e^xu=ex and du=ex dxdu = e^x \, dxdu=exdx into the integral:∫01ex1+e2x dx=∫1e11+u2 du\int_0^1 \frac{e^x}{1 + e^{2x}} \, dx = \int_1^e \frac{1}{1 + u^2} \, du∫01​1+e2xex​dx=∫1e​1+u21​du

Step 3: Recognizing the standard integral

The integral:∫11+u2 du\int \frac{1}{1 + u^2} \, du∫1+u21​du

is a standard form, and its solution is:arctan⁡(u)\arctan(u)arctan(u)

Thus, our integral becomes:∫1e11+u2 du=arctan⁡(u)∣1e\int_1^e \frac{1}{1 + u^2} \, du = \arctan(u) \bigg|_1^e∫1e​1+u21​du=arctan(u)​1e​

Step 4: Evaluate the definite integral

Now, evaluate the arctangent at the bounds u=eu = eu=e and u=1u = 1u=1:arctan⁡(e)−arctan⁡(1)\arctan(e) – \arctan(1)arctan(e)−arctan(1)

We know that:arctan⁡(1)=π4\arctan(1) = \frac{\pi}{4}arctan(1)=4π​

Thus, the integral becomes:arctan⁡(e)−π4\arctan(e) – \frac{\pi}{4}arctan(e)−4π​

Final Answer

The value of the definite integral is:arctan⁡(e)−π4\arctan(e) – \frac{\pi}{4}arctan(e)−4π​

This is the exact answer for the given definite integral.