How do you integrate $\int \frac{e^{2 x}}{1 + e^{2 x}} d x$ $int e^(2x)/(1+e^(2x))dx$ ?

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Answer 1 · 2016-11-21T06:00:20

The answer is $= \frac{1}{2} ln (1 + e^{2 x}) + C$ $=1/2ln(1+e^(2x))+C$

Let's do it by substitution

Let $u = 1 + e^{2 x}$ $u=1+e^(2x)$

then, $d u = 2 e^{2 x} d x$ $du=2e^(2x)dx$

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

$= \frac{1}{2} \int \frac{d u}{u} = \frac{1}{2} ln | u |$ $=1/2int(du)/u=1/2lnabsu$

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

How do you integrate ∫e2x1+e2xdxint e^(2x)/(1+e^(2x))dx?