Question

24 multiple choice 1 point evaluate. \\(\int \frac{7e^{3x}}{e^{3x} + 1} dx\\) \\(\frac{7}{3}(e^{3x} + 1)^{-2} + c\\) \\(\frac{7}{3} \ln (e^{3x} + 1) + c\\) \\(\frac{3}{7} \ln (e^{3x} + 1) + c\\) \\(7 \ln (e^{3x} + 1) + c\\)

Explanation:

Step1: Choose substitution

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

Step2: Rewrite the integral

The integral is \( \int\frac{7e^{3x}}{e^{3x}+1}dx \). Substitute \( u \) and \( e^{3x}dx=\frac{du}{3} \):
\( \int\frac{7\cdot\frac{du}{3}}{u}=\frac{7}{3}\int\frac{1}{u}du \)

Step3: Integrate \( \frac{1}{u} \)

The integral of \( \frac{1}{u} \) with respect to \( u \) is \( \ln|u|+C \). So:
\( \frac{7}{3}\ln|u|+C \)

Step4: Substitute back \( u \)

Since \( u = e^{3x}+1 \) (and \( e^{3x}+1>0 \) for all real \( x \), we can drop the absolute value):
\( \frac{7}{3}\ln(e^{3x}+1)+C \)

Answer:

\(\frac{7}{3} \ln (e^{3x} + 1) + C\) (the second option)