Exercise

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

Step-by-step Solution

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 finding the derivative of the equation above

$du=2dx$
3

Isolate $dx$ in the previous equation

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

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

$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
8

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

$\begin{matrix}\displaystyle{u=u}\\ \displaystyle{du=du}\end{matrix}$
9

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=e^udu}\\ \displaystyle{\int dv=\int e^udu}\end{matrix}$
10

Solve the integral to find $v$

$v=\int e^udu$
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 $a \neq 1$

$e^u$
12

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

$\frac{1}{4}\left(e^u\cdot u-\int e^udu\right)$
13

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

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

Multiplying the fraction by $-1$

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

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

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

Multiply the fraction and term in $2\cdot \frac{1}{4}e^{2x}x$

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

The integral $-\frac{1}{4}\int e^udu$ 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 to the exercise

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

Try other ways to solve this exercise

  • Integrate by substitution
  • Integrate by partial fractions
  • Integrate by parts
  • 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...
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