What is the antiderivative of $e^{2 x}$ $e^(2x)$ ?

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Answer 1 · 2016-02-13T21:35:01

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

The antiderivative of $e^{2 x}$ $e^(2x)$ is equivalent to

$= \int e^{2 x} d x$ $=inte^(2x)dx$

Let $u = 2 x$ $u=2x$ , so $d u = 2 d x$ $du=2dx$ .

Here, we can make some substitutions:

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

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

The $\frac{1}{2}$ $1/2$ can be brought out of the integral:

$= \frac{1}{2} \int e^{u} d u$ $=1/2inte^udu$

Note that $\int e^{u} d u = e^{u} + C$ $inte^udu=e^u+C$ , so

$= \frac{e^{u}}{2} + C$ $=e^u/2+C$

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

What is the antiderivative of e2xe^(2x) ?