What is the integral of $e^{3 x}$ $e^(3x)$ ?

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Answer 1 · 2015-05-19T21:02:15

The answer is $\int e^{3 x} d x = \frac{e^{3 x}}{3}$ $inte^(3x)dx=e^(3x)/3$ .

So we have $f (x) = e^{3 x} = g (h (x))$ $f(x) = e^(3x) = g(h(x))$ , where $g (x) = e^{x}$ $g(x) = e^x$ and $h (x) = 3 x$ $h(x) = 3x$ .

The antiderivative of such a form is given by :

$intg(h(x))*h'(x)dx = G(h(x))$

We know that the derivative of $h(x) = 3x$ is $h'(x)=3$ .

We also know that the antiderivative of $g(x) = e^x$ is $G(x) = e^x$ .

We have $inte^(3x)dx$ but, with our formula, we can only calculate $inte^(3x)*3dx$ , so what we will do is :

$inte^(3x)dx = 1/3inte^(3x)*3dx = e^(3x)/3$ .

That's it.

What is the integral of e^(3x)e3xe^(3x)?