How do you find the antiderivative of $((2 x) e^{3 x})$ $((2x)e^(3x))$ ?

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How do you find the antiderivative of $((2 x) e^{3 x})$ $((2x)e^(3x))$ ?

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Answer 1 · 2016-08-03T13:18:44

$\frac{1}{9} \cdot (3 x - 1) \cdot e^{3 x} + C$ $1/9*(3x-1)*e^(3x)+C$ .

Explanation:

Let, $I = \int 2 x e^{3 x} d x \Rightarrow I = 2 \int x e^{3 x} d x$ $I=int2xe^(3x)dx rArr I=2intxe^(3x)dx$ .

To find $I$ $I$ , we will use the following Rule of Integration by Parts :

$\int u v d x = u \int v d x - \int {\frac{d u}{d x} \int v d x} d x$ $intuvdx=uintvdx-int{(du)/dxintvdx}dx$ .

We take, $u = x, s o, \frac{d u}{d x} = 1, &, v = e^{3 x}, s o, \int v d x = \frac{1}{3} e^{3 x}$ $u=x, so, (du)/dx=1, &, v=e^(3x), so, intvdx=1/3e^(3x)$ . So,

$I = x \cdot \frac{1}{3} e^{3 x} - \int {1 \cdot \frac{1}{3} e^{3 x}} d x$ $I=x*1/3e^(3x)-int{1*1/3e^(3x)}dx$

$= \frac{x}{3} e^{3 x} - \frac{1}{3} \int e^{3 x} d x$ $=x/3e^(3x)-1/3inte^(3x)dx$

$= \frac{x}{3} e^{3 x} - \frac{1}{3} \cdot \frac{1}{3} e^{3 x}$ $=x/3e^(3x)-1/3*1/3e^(3x)$

$:. I = 1/9*(3x-1)*e^(3x)+C$ .

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How do you find the antiderivative of $((2 x) e^{3 x})$ $((2x)e^(3x))$ ?

1 Answer

Explanation:

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