Step-by-step Solution

Find the integral $\int xe^{2x}dx$

Go!
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Final Answer

$\frac{1}{2}e^{2x}x-\frac{1}{4}e^{2x}+C_0$
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Step-by-step Solution

Problem to solve:

$\int\left(x\cdot e^{2x}\right)dx$

Solving method

1

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 $1$

$1$
2

First, identify $u$ and calculate $du$

$\begin{matrix}\displaystyle{u=x}\\ \displaystyle{du=dx}\end{matrix}$
3

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=e^{2x}dx}\\ \displaystyle{\int dv=\int e^{2x}dx}\end{matrix}$
4

Solve the integral

$v=\int e^{2x}dx$
5

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

$2$
6

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

$du=2dx$
7

Isolate $dx$ in the previous equation

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

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

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

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

Divide $1$ by $2$

$\frac{1}{2}\int e^udu$
9

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

$\frac{1}{2}\int e^udu$
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 $a \neq 1$

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

$\frac{1}{2}e^{2x}$
11

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

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

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

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

The integral of a constant by a function is equal to the constant multiplied by the integral of the function

$\frac{1}{2}e^{2x}x-\frac{1}{2}\int e^{2x}dx$
12

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

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

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

$2$
14

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

$du=2dx$
15

Isolate $dx$ in the previous equation

$\frac{du}{2}=dx$
16

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

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

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

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

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: $2x$

$-\frac{1}{4}e^{2x}$
17

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

$-\frac{1}{4}e^{2x}$
18

Gather the results of all integrals

$\frac{1}{2}e^{2x}x-\frac{1}{4}e^{2x}$
19

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

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

Final Answer

$\frac{1}{2}e^{2x}x-\frac{1}{4}e^{2x}+C_0$
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0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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