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: $$\frac{1}{3} \ln (e^{3x} + 1) + C$$