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

Step-by-step Solution

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Final answer to the problem

$\frac{1}{2}e^{2x}x-\frac{1}{4}e^{2x}+C_0$
Got another answer? Verify it here!

Step-by-step Solution

How should I solve this problem?

  • Integrate by parts
  • Integrate by partial fractions
  • Integrate by substitution
  • Integrate using tabular integration
  • Integrate by trigonometric substitution
  • Weierstrass Substitution
  • Integrate using trigonometric identities
  • Integrate using basic integrals
  • Product of Binomials with Common Term
  • FOIL Method
  • Load more...
Can't find a method? Tell us so we can add it.
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 or choose $u$ and calculate it's derivative, $du$

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

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

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

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

$2$
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

$du=2dx$

Rearrange the equation

$2dx=du$

Divide both sides of the equation by $2$

$dx=\frac{du}{2}$
7

Isolate $dx$ in the previous equation

$dx=\frac{du}{2}$
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Substituting $u$ and $dx$ in the integral and simplify

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

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

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

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

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

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

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 $-1$

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

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

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

$2$
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

$du=2dx$

Rearrange the equation

$2dx=du$

Divide both sides of the equation by $2$

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

Isolate $dx$ in the previous equation

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

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

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: $2x$

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

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

$-\frac{1}{4}e^{2x}$
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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 to the problem

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

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Function Plot

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

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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

How to improve your answer:

Main Topic: Integrals of Exponential Functions

Those are integrals that involve exponential functions. Recall that an exponential function is a function of the form f(x)=a^x.

Used Formulas

See formulas (5)

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