Integral Exercise 30

We are asked to evaluate the following integral:

$$ \int \frac{1}{1 + e^{3x}} \ dx $$

To simplify the expression, we multiply and divide the integrand by \( e^{-3x} \):

$$ \int \frac{1}{1 + e^{3x}} \cdot \frac{e^{-3x}}{e^{-3x}} \ dx $$

This gives:

$$ \int \frac{e^{-3x}}{(1 + e^{3x}) \cdot e^{-3x}} \ dx = \int \frac{e^{-3x}}{e^{-3x} + 1} \ dx $$

We now perform a substitution. Let:

$$ u = e^{-3x} + 1 $$

Differentiating both sides with respect to \( x \), we find:

$$ du = -3e^{-3x} \ dx \quad \Rightarrow \quad \frac{du}{-3} = e^{-3x} \ dx $$

Substituting into the integral, we obtain:

$$ \int \frac{1}{u} \cdot \frac{du}{-3} = -\frac{1}{3} \int \frac{1}{u} \ du $$

This is a standard logarithmic integral:

$$ -\frac{1}{3} \ln |u| + c $$

Substituting back \( u = e^{-3x} + 1 \), we conclude that:

$$ \int \frac{1}{1 + e^{3x}} \ dx = -\frac{1}{3} \ln |e^{-3x} + 1| + c $$

And that completes the evaluation.