Question

determine the integral by making the appropriate substitution.

\\( \int \frac { e ^ { 4 x } } { 3 + 2 e ^ { 4 x } } d x \\)

\\( \int \frac { e ^ { 4 x } } { 3 + 2 e ^ { 4 x } } d x = \square \\)
(use parentheses to clearly denote the argument of each function.)

Explanation:

Step1: Set substitution

Let \(u = 3 + 2e^{4x}\). Then, find \(du\).
Differentiate \(u\) with respect to \(x\): \(du=2e^{4x}\times4dx = 8e^{4x}dx\), so \(e^{4x}dx=\frac{1}{8}du\).

Step2: Substitute into integral

The integral \(\int\frac{e^{4x}}{3 + 2e^{4x}}dx\) becomes \(\int\frac{1}{u}\times\frac{1}{8}du\).

Step3: Integrate

\(\frac{1}{8}\int\frac{1}{u}du=\frac{1}{8}\ln|u|+C\).

Step4: Substitute back

Substitute \(u = 3 + 2e^{4x}\) back in. The integral is \(\frac{1}{8}\ln(3 + 2e^{4x})+C\).

Answer:

\(\frac{1}{8}\ln(3 + 2e^{4x})+C\)