Question

Integrate the following function w.r.t. $x$: $\int \frac{e^{3x}}{e^{3x} + 1} \, dx$

• A. \( \frac{1}{3} \ln \left( e^{3x} + 1 \right) + C \)
• B. \( \frac{1}{3} \ln \left( e^{3x} - 1 \right) + C \)
• C. \( \frac{1}{3} \ln \left( e^{3x} + e^x \right) + C \)
• D. \( \frac{1}{2} \ln \left( e^{3x} + 1 \right) + C \)

Hint:

To simplify integrals with exponential functions, use substitution to transform the integral into a standard form.

Answer 1

The Correct Option is A

Step 1: Observe the integral and try substitution. Let: \[ u = e^{3x} + 1 \quad \Rightarrow \quad du = 3e^{3x} dx \] \[ \text{Thus,} \quad \frac{e^{3x}}{e^{3x} + 1} dx = \frac{1}{3} \cdot \frac{du}{u} \] Step 2: Now, the integral becomes: \[ \int \frac{1}{3} \cdot \frac{du}{u} = \frac{1}{3} \ln |u| + C \] \[ \text{Substitute back } u = e^{3x} + 1: \] \[ \frac{1}{3} \ln |e^{3x} + 1| + C \] Thus, the solution is: \[ \frac{1}{3} \ln (e^{3x} + 1) + C \]