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

Step-by-step Solution

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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

Specify the solving method

1

We can solve the integral $\int xe^{2x}dx$ 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

$u=2x$
2

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

$du=2dx$
3

Isolate $dx$ in the previous equation

$du=2dx$
4

Rewriting $x$ in terms of $u$

$x=\frac{u}{2}$
5

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

$\int\frac{ue^u}{4}du$
6

Take the constant $\frac{1}{4}$ out of the integral

$\frac{1}{4}\int ue^udu$
7

Divide $1$ by $4$

$\frac{1}{4}\int ue^udu$
8

We can solve the integral $\int ue^udu$ by applying integration by parts method to calculate the integral of the product of two functions, using the following formula

$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
9

First, identify $u$ and calculate $du$

$\begin{matrix}\displaystyle{u=u}\\ \displaystyle{du=du}\end{matrix}$
10

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=e^udu}\\ \displaystyle{\int dv=\int e^udu}\end{matrix}$
11

Solve the integral

$v=\int e^udu$
12

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 $a \neq 1$

$e^u$
13

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

$\frac{1}{4}\left(e^u\cdot u-\int e^udu\right)$
14

Multiply the single term $\frac{1}{4}$ by each term of the polynomial $\left(e^u\cdot u-\int e^udu\right)$

$\frac{1}{4}e^u\cdot u-\frac{1}{4}\int e^udu$
15

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

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

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

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

Gather the results of all integrals

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

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

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

Explore different ways to solve this problem

Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more

Solve integral of xe^2xdx using partial fractionsSolve integral of xe^2xdx using basic integralsSolve integral of xe^2xdx using integration by partsSolve integral of xe^2xdx using tabular integrationSolve integral of xe^2xdx using trigonometric substitution

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

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1
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3
4
5
6
7
8
9
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

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