# Integrate xe^(2x)*5

## \int5x e^{2x}dx

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$-\frac{5}{4}e^{2x}+\frac{5}{2}xe^{2x}+C_0$

## Step by step solution

Problem

$\int5x e^{2x}dx$
1

Taking the constant out of the integral

$5\int xe^{2x}dx$
2

Use the integration by parts theorem to calculate the integral $\int xe^{2x}dx$, using the following formula

$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
3

First, identify $u$ and calculate $du$

$\begin{matrix}\displaystyle{u=x}\\ \displaystyle{du=dx}\end{matrix}$
4

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=e^{2x}dx}\\ \displaystyle{\int dv=\int e^{2x}dx}\end{matrix}$
5

Solve the integral

$v=\int e^{2x}dx$
6

Solve the integral $\int e^{2x}dx$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=2x \\ du=2dx\end{matrix}$
7

Isolate $dx$ in the previous equation

$\frac{du}{2}=dx$
8

Substituting $u$ and $dx$ in the integral

$5\int x\frac{e^u}{2}dx$
9

Taking the constant out of the integral

$5\int x\frac{e^u}{2}dx$
10

The integral of the exponential function is given by the following formula $\displaystyle \int a^xdx=\frac{a^x}{\ln(a)}$, where $a > 0$ and $a \neq 1$

$5\int x\frac{e^u}{2}dx$
11

Substitute $u$ back for it's value, $2x$

$5\int x\frac{e^u}{2}dx$
12

Now replace the values of $u$, $du$ and $v$ in the last formula

$5\left(\frac{1}{2}xe^{2x}-\frac{1}{2}\int e^{2x}dx\right)$
13

Solve the integral $\int e^{2x}dx$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=2x \\ du=2dx\end{matrix}$
14

Isolate $dx$ in the previous equation

$\frac{du}{2}=dx$
15

Substituting $u$ and $dx$ in the integral

$5\left(\frac{1}{2}xe^{2x}-\frac{1}{2}\int\frac{e^u}{2}du\right)$
16

Taking the constant out of the integral

$5\left(\frac{1}{2}xe^{2x}-\frac{1}{2}\cdot \frac{1}{2}\int e^udu\right)$
17

Multiply $\frac{1}{2}$ times $-\frac{1}{2}$

$5\left(\frac{1}{2}xe^{2x}-\frac{1}{4}\int e^udu\right)$
18

The integral of the exponential function is given by the following formula $\displaystyle \int a^xdx=\frac{a^x}{\ln(a)}$, where $a > 0$ and $a \neq 1$

$5\left(\frac{1}{2}xe^{2x}-\frac{1}{4}e^u\right)$
19

Substitute $u$ back for it's value, $2x$

$5\left(\frac{1}{2}xe^{2x}-\frac{1}{4}e^{2x}\right)$
20

Multiply $\left(\frac{1}{2}xe^{2x}+-\frac{1}{4}e^{2x}\right)$ by $5$

$\frac{5}{2}xe^{2x}-\frac{5}{4}e^{2x}$
21

$-\frac{5}{4}e^{2x}+\frac{5}{2}xe^{2x}+C_0$

$-\frac{5}{4}e^{2x}+\frac{5}{2}xe^{2x}+C_0$

### Main topic:

Integration by parts

0.27 seconds

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