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

Step-by-step Solution

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Final answer to the problem

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

How should I solve this problem?

  • Integrate using tabular integration
  • Integrate by partial fractions
  • Integrate by substitution
  • Integrate by parts
  • Integrate by trigonometric substitution
  • Weierstrass Substitution
  • Integrate using trigonometric identities
  • Integrate using basic integrals
  • Product of Binomials with Common Term
  • FOIL Method
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1

We can solve the integral $\int xe^{2x}dx$ by applying the method of tabular integration by parts, which allows us to perform successive integrations by parts on integrals of the form $\int P(x)T(x) dx$. $P(x)$ is typically a polynomial function and $T(x)$ is a transcendent function such as $\sin(x)$, $\cos(x)$ and $e^x$. The first step is to choose functions $P(x)$ and $T(x)$

$\begin{matrix}P(x)=x \\ T(x)=e^{2x}\end{matrix}$

Find the derivative of $x$ with respect to $x$

$x$

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

$1$

The derivative of the constant function ($1$) is equal to zero

0
2

Derive $P(x)$ until it becomes $0$

$0$

Find the integral of $e^{2x}$ with respect to $x$

$e^{2x}$

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

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

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

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

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 $a \neq 1$

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

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

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

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

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

When multiplying two powers that have the same base ($\frac{1}{2}$), you can add the exponents

$\left(\frac{1}{2}\right)^2\int e^udu$

Calculate the power $\left(\frac{1}{2}\right)^2$

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

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: $2x$

$\frac{1}{4}e^{2x}$
3

Integrate $T(x)$ as many times as we have had to derive $P(x)$, so we must integrate $e^{2x}$ a total of $2$ times

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

With the derivatives and integrals of both functions we build the following table

$\begin{matrix}\mathrm{Derivatives} & \mathrm{Sign} & \mathrm{Integrals} \\ & & e^{2x} \\ x & + & \frac{1}{2}e^{2x} \\ 1 & - & \frac{1}{4}e^{2x} \\ 0 & & \end{matrix}$
5

Then the solution is the sum of the products of the derivatives and the integrals according to the previous table. The first term consists of the product of the polynomial function by the first integral. The second term is the product of the first derivative by the second integral, and so on.

$\frac{1}{2}xe^{2x}- \left(\frac{1}{4}\right)e^{2x}$
6

Multiply the fraction and term in $- \left(\frac{1}{4}\right)e^{2x}$

$\frac{1}{2}xe^{2x}-\frac{1}{4}e^{2x}$
7

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

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

Final answer to the problem

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

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

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0
a
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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 (3)

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