Integral Exercise
Let's evaluate the following integral:
$$ \int \frac{3e^{3x}}{6-e^{3x}} \, dx $$
To tackle this, we'll apply the substitution method. The idea is to simplify the integrand by introducing a new variable that replaces part of the expression.
A suitable substitution in this case is u = 6 - e3x.
$$ u = 6 - e^{3x} $$
We now differentiate \( u \) with respect to \( x \) to find \( du \), and express \( dx \) in terms of \( du \):
$$
\begin{align*}
u &= 6 - e^{3x} \\
\frac{du}{dx} &= -3e^{3x} \\
dx &= -\frac{1}{3e^{3x}} \, du
\end{align*} $$
Now we rewrite the original integral using this substitution:
$$ \int \frac{3e^{3x}}{6-e^{3x}} \, dx $$
Substituting dx = -\frac{1}{3e3x}\, du, we get:
$$ \int \frac{3e^{3x}}{6-e^{3x}} \cdot \left( -\frac{1}{3e^{3x}} \, du \right) $$
Next, replace 6 - e3x with \( u \):
$$ \int \frac{3e^{3x}}{u} \cdot \left( -\frac{1}{3e^{3x}} \, du \right) $$
We simplify by canceling out the exponential terms:
$$ \int - \frac{1}{u} \, du $$
The constant factor can be pulled out using the linearity property of integration:
$$ -\int \frac{1}{u} \, du $$
We’ve now reduced the problem to a basic integral.
The antiderivative of \( -\frac{1}{u} \) is simply \( -\ln|u| + C \), where \( C \) is the constant of integration:
$$ -\int \frac{1}{u} \, du = -\ln|u| + C $$
Substituting back \( u = 6 - e^{3x} \), we get:
$$ -\ln|6 - e^{3x}| + C $$
So the final result is the natural logarithm of the absolute value of \( 6 - e^{3x} \):
$$ \int \frac{3e^{3x}}{6-e^{3x}} \, dx = -\ln|6 - e^{3x}| + C $$
And that completes the solution.
