How do you find the antiderivative of $[\frac{e^{2 x}}{4 + e^{4 x}}]$ $[e^(2x)/(4+e^(4x))]$ ?

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Answer 1 · 2016-08-03T14:19:01

$\frac{1}{4} arctan (\frac{e^{2 x}}{2}) + C$ $1/4arctan(e^(2x)/2)+C$ .

Let, $I = \int \frac{e^{2 x}}{4 + e^{4 x}} d x$ $I = inte^(2x)/(4+e^(4x))dx$ .

We take subst. $e^{2 x} = t \Rightarrow e^{2 x} \cdot 2 d x = d t$ $e^(2x)=t rArr e^(2x)*2dx=dt$ .

Therefore, $I = \frac{1}{2} \int \frac{2 \cdot e^{2 x} \cdot d x}{4 + {(e^{2 x})}^{2}}$ $I=1/2int(2*e^(2x)*dx)/{4+(e^(2x))^2}$ ,

$= \frac{1}{2} \int \frac{d t}{4 + t^{2}} = \frac{1}{2} \cdot \frac{1}{2} \cdot arctan (\frac{t}{2})$ $=1/2intdt/(4+t^2) = 1/2*1/2*arctan(t/2)$ .

Hence, $I = \frac{1}{4} arctan (\frac{e^{2 x}}{2}) + C$ $I=1/4arctan(e^(2x)/2)+C$ .

How do you find the antiderivative of [e^(2x)/(4+e^(4x))][e2x4+e4x][e^(2x)/(4+e^(4x))]?