The problem shown in the image asks how the given integral changes after making the substitution $e^{2x} = t$.

The original integral is:

$\int_{-\frac{1}{2} \ln 3}^{0} \frac{dx}{3 + e^{2x}}$

Step-by-Step Solution:

1. Substitution: $e^{2x} = t$ Differentiating this with respect to $x$: $2e^{2x}dx = dt \quad \text{or} \quad dx = \frac{dt}{2t}$

2. Changing the limits of integration:

   For $x = 0$: $e^{2(0)} = 1 \quad \Rightarrow \quad t = 1$

   For $x = -\frac{1}{2} \ln 3$: $e^{2 \left(-\frac{1}{2} \ln 3\right)} = e^{-\ln 3} = \frac{1}{3} \quad \Rightarrow \quad t = \frac{1}{3}$

3. Substituting into the integral:

   The integral becomes: $\int_{\frac{1}{3}}^1 \frac{\frac{dt}{2t}}{3 + t}$ This simplifies to: $\frac{1}{2} \int_{\frac{1}{3}}^1 \frac{dt}{t(3 + t)}$

Thus, the integral after substitution takes the form: $\frac{1}{2} \int_{\frac{1}{3}}^1 \frac{dt}{t(3+t)}$

This matches the third option in the list.

Answer:

The correct option is: $\frac{1}{2} \int_{\frac{1}{3}}^1 \frac{dt}{t(3+t)}$

Let me know if you would like more details, and here are some related questions for further practice:

1. How do you compute the integral $\int \frac{dx}{e^x + 1}$?
2. What happens when you substitute $e^x = t$ in an integral with a different exponent?
3. How do you handle integrals with logarithmic limits like $\int_{-\ln a}^{\ln b} f(x) dx$?
4. What are common techniques for solving integrals involving exponential functions?
5. How do you find the derivative of functions involving natural logarithms?

Tip: Always remember to adjust the limits of integration when making substitutions!