How do you evaluate the integral $\int \frac{2^{x}}{2^{x} + 1}$ $int 2^x/(2^x+1)$ ?

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How do you evaluate the integral $\int \frac{2^{x}}{2^{x} + 1}$ $int 2^x/(2^x+1)$ ?

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Answer 1 · 2018-03-23T02:40:42

$\frac{ln (2^{x} + 1)}{ln 2} + C$ $ln(2^x+1)/ln 2+C$

Explanation:

Since $\frac{d}{d x} (2^{x} + 1) = 2^{x} ln 2$ $d/dx(2^x+1)=2^x ln 2$ , we have

$\int \frac{2^{x}}{2^{x} + 1} d x = \frac{1}{ln 2} \int \frac{d (2^{x} + 1)}{2^{x} + 1} = \frac{ln (2^{x} + 1)}{ln 2} + C$ $int 2^x/(2^x+1)dx = 1/ln 2 int {d(2^x+1)}/(2^x+1) = ln(2^x+1)/ln 2+C$

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How do you evaluate the integral $\int \frac{2^{x}}{2^{x} + 1}$ $int 2^x/(2^x+1)$ ?

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How do you evaluate the integral $\int \frac{2^{x}}{2^{x} + 1}$ $int 2^x/(2^x+1)$ ?