Single Variable Calculus, Chapter 7, 7.4-2, Section 7.4-2, Problem 50

Find the integral $\displaystyle \int\frac{2^x}{2^x + 1} dx$

If we let $u = 2^x + 1$, then $du = 2^x \ln 2 dx$, so $\displaystyle 2^x dx = \frac{du}{\ln 2}$. Thus,


$
\begin{equation}
\begin{aligned}

\int \frac{2^x}{2^x + 1} dx =& \int \frac{1}{2^x + 1} 2^x dx
\\
\\
\int \frac{2^x}{2^x + 1} dx =& \int \frac{1}{u} \frac{du}{\ln 2}
\\
\\
\int \frac{2^x}{2^x + 1} dx =& \frac{1}{\ln 2} \int \frac{1}{u} du
\\
\\
\int \frac{2^x}{2^x + 1} dx =& \frac{1}{\ln 2} \cdot \ln |u| + C
\\
\\
\int \frac{2^x}{2^x + 1} dx =& \frac{1}{\ln 2} \cdot \ln (2^x + 1) + C
\\
\\
\int \frac{2^x}{2^x + 1} dx =& \frac{\ln (2^x + 1)}{\ln 2} + C

\end{aligned}
\end{equation}
$