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1

Here, we show you a step-by-step solved example of calculus. This solution was automatically generated by our smart calculator:

$\int x e^{2x}dx$
2

We can solve the integral $\int xe^{2x}dx$ by applying integration by parts method to calculate the integral of the product of two functions, using the following formula

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

The derivative of the linear function is equal to $1$

$1$
3

First, identify or choose $u$ and calculate it's derivative, $du$

$\begin{matrix}\displaystyle{u=x}\\ \displaystyle{du=dx}\end{matrix}$
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 to find $v$

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

We can solve the integral $\int e^{2x}dx$ by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it $u$), which when substituted makes the integral easier. We see that $2x$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=2x$

Differentiate both sides of the equation $u=2x$

$du=\frac{d}{dx}\left(2x\right)$

Find the derivative

$\frac{d}{dx}\left(2x\right)$

The derivative of the linear function times a constant, is equal to the constant

$2\frac{d}{dx}\left(x\right)$

The derivative of the linear function is equal to $1$

$2$
7

Now, in order to rewrite $dx$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above

$du=2dx$

Rearrange the equation

$2dx=du$

Divide both sides of the equation by $2$

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

Isolate $dx$ in the previous equation

$dx=\frac{du}{2}$
9

Substituting $u$ and $dx$ in the integral and simplify

$\int\frac{e^u}{2}du$
10

Take the constant $\frac{1}{2}$ out of the integral

$\frac{1}{2}\int e^udu$
11

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 $a \neq 1$

$\frac{1}{2}e^u$

Replace $u$ with the value that we assigned to it in the beginning: $2x$

$\frac{1}{2}e^{2x}$
12

Replace $u$ with the value that we assigned to it in the beginning: $2x$

$\frac{1}{2}e^{2x}$

The integral of a function times a constant ($\frac{1}{2}$) is equal to the constant times the integral of the function

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

Multiplying the fraction by $-1$

$\frac{1}{2}e^{2x}x-\frac{1}{2}\int e^{2x}dx$
13

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

$\frac{1}{2}e^{2x}x-\frac{1}{2}\int e^{2x}dx$

We can solve the integral $\int e^{2x}dx$ by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it $u$), which when substituted makes the integral easier. We see that $2x$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=2x$

Now, in order to rewrite $dx$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above

$du=2dx$

Isolate $dx$ in the previous equation

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

Substituting $u$ and $dx$ in the integral and simplify

$-\frac{1}{2}\int\frac{e^u}{2}du$

Take the constant $\frac{1}{2}$ out of the integral

$-\frac{1}{2}\cdot \frac{1}{2}\int e^udu$

Multiplying fractions $-\frac{1}{2} \times \frac{1}{2}$

$-\frac{1}{4}\int e^udu$

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 $a \neq 1$

$-\frac{1}{4}e^u$

Replace $u$ with the value that we assigned to it in the beginning: $2x$

$-\frac{1}{4}e^{2x}$
14

The integral $-\frac{1}{2}\int e^{2x}dx$ results in: $-\frac{1}{4}e^{2x}$

$-\frac{1}{4}e^{2x}$
15

Gather the results of all integrals

$\frac{1}{2}e^{2x}x-\frac{1}{4}e^{2x}$
16

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

$\frac{1}{2}e^{2x}x-\frac{1}{4}e^{2x}+C_0$

Endgültige Antwort auf das Problem

$\frac{1}{2}e^{2x}x-\frac{1}{4}e^{2x}+C_0$

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