How do you find the antiderivative of $\frac{e^{2 x}}{1 + (e^{4 x}) d x}$ $(e^(2x))/(1+(e^(4x))dx$ ?

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How do you find the antiderivative of $\frac{e^{2 x}}{1 + (e^{4 x}) d x}$ $(e^(2x))/(1+(e^(4x))dx$ ?

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Answer 1 · 2016-12-12T20:46:54

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

Explanation:

$\int \frac{e^{2 x}}{1 + e^{4 x}} d x$ $inte^(2x)/(1+e^(4x))dx$

Let $u = e^{2 x}$ $u=e^(2x)$ so $d u = 2 e^{2 x} d x$ $du=2e^(2x)dx$ .

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

$= \frac{1}{2} \int \frac{1}{1 + u^{2}} d u$ $=1/2int1/(1+u^2)du$

This is the arctangent integral:

$= \frac{1}{2} arctan (u) + C$ $=1/2arctan(u)+C$

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

Another way to show this is to use the trigonometric substitution $e^{2 x} = tan (θ)$ $e^(2x)=tan(theta)$ .

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How do you find the antiderivative of $\frac{e^{2 x}}{1 + (e^{4 x}) d x}$ $(e^(2x))/(1+(e^(4x))dx$ ?

1 Answer

Explanation:

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How do you find the antiderivative of $\frac{e^{2 x}}{1 + (e^{4 x}) d x}$ $(e^(2x))/(1+(e^(4x))dx$ ?