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Integral of d{x}:
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Integral of xe^-1
Integral of x^3*ln(2x)
Identical expressions
x^ three *ln(2x)
x cubed multiply by ln(2x)
x to the power of three multiply by ln(2x)
x3*ln(2x)
x3*ln2x
x³*ln(2x)
x to the power of 3*ln(2x)
x^3ln(2x)
x3ln(2x)
x3ln2x
x^3ln2x
x^3*ln(2x)dx

Solving integrals/ x^3*ln(2x)

Integral of x^3*ln(2x) dx

The solution

You have entered [src]

  1               
  /               
 |                
 |   3            
 |  x *log(2*x) dx
 |                
/                 
0

$$\int\limits_{0}^{1} x^{3} \log{\left(2 x \right)}\, dx$$

Integral(x^3*log(2*x), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $x^{3} \log{\left(2 x \right)} = x^{3} \log{\left(x \right)} + x^{3} \log{\left(2 \right)}$
2. Integrate term-by-term:
  1. Let $u = \log{\left(x \right)}$ .
    
    Then let $du = \frac{dx}{x}$ and substitute $du$ :
    
    $\int u e^{4 u}\, du$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u$ and let $\operatorname{dv}{\left(u \right)} = e^{4 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = 4 u$ .
      
      Then let $du = 4 du$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{e^{u}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{4 u}}{4}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{4 u}}{4}\, du = \frac{\int e^{4 u}\, du}{4}$
      
      Let $u = 4 u$ .
      
      Then let $du = 4 du$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{e^{u}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{4 u}}{4}$
      
      So, the result is: $\frac{e^{4 u}}{16}$
    Now substitute $u$ back in:
    
    $\frac{x^{4} \log{\left(x \right)}}{4} - \frac{x^{4}}{16}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int x^{3} \log{\left(2 \right)}\, dx = \log{\left(2 \right)} \int x^{3}\, dx$
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x^{3}\, dx = \frac{x^{4}}{4}$
    So, the result is: $\frac{x^{4} \log{\left(2 \right)}}{4}$
  The result is: $\frac{x^{4} \log{\left(x \right)}}{4} - \frac{x^{4}}{16} + \frac{x^{4} \log{\left(2 \right)}}{4}$
Method #2
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = \log{\left(2 x \right)}$ and let $\operatorname{dv}{\left(x \right)} = x^{3}$ .
  
  Then $\operatorname{du}{\left(x \right)} = \frac{1}{x}$ .
  
  To find $v{\left(x \right)}$ :
  1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
    
    $\int x^{3}\, dx = \frac{x^{4}}{4}$
  Now evaluate the sub-integral.
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{x^{3}}{4}\, dx = \frac{\int x^{3}\, dx}{4}$
  1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
    
    $\int x^{3}\, dx = \frac{x^{4}}{4}$
  So, the result is: $\frac{x^{4}}{16}$
Method #3
1. Rewrite the integrand:
  
  $x^{3} \log{\left(2 x \right)} = x^{3} \log{\left(x \right)} + x^{3} \log{\left(2 \right)}$
2. Integrate term-by-term:
  1. Let $u = \log{\left(x \right)}$ .
    
    Then let $du = \frac{dx}{x}$ and substitute $du$ :
    
    $\int u e^{4 u}\, du$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u$ and let $\operatorname{dv}{\left(u \right)} = e^{4 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = 4 u$ .
      
      Then let $du = 4 du$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{e^{u}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{4 u}}{4}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{4 u}}{4}\, du = \frac{\int e^{4 u}\, du}{4}$
      
      Let $u = 4 u$ .
      
      Then let $du = 4 du$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{e^{u}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{4 u}}{4}$
      
      So, the result is: $\frac{e^{4 u}}{16}$
    Now substitute $u$ back in:
    
    $\frac{x^{4} \log{\left(x \right)}}{4} - \frac{x^{4}}{16}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int x^{3} \log{\left(2 \right)}\, dx = \log{\left(2 \right)} \int x^{3}\, dx$
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x^{3}\, dx = \frac{x^{4}}{4}$
    So, the result is: $\frac{x^{4} \log{\left(2 \right)}}{4}$
  The result is: $\frac{x^{4} \log{\left(x \right)}}{4} - \frac{x^{4}}{16} + \frac{x^{4} \log{\left(2 \right)}}{4}$
Now simplify:

$\frac{x^{4} \left(4 \log{\left(x \right)} - 1 + \log{\left(16 \right)}\right)}{16}$
Add the constant of integration:

$\frac{x^{4} \left(4 \log{\left(x \right)} - 1 + \log{\left(16 \right)}\right)}{16}+ \mathrm{constant}$

The answer is:

$\frac{x^{4} \left(4 \log{\left(x \right)} - 1 + \log{\left(16 \right)}\right)}{16}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                               
 |                       4    4           4       
 |  3                   x    x *log(2)   x *log(x)
 | x *log(2*x) dx = C - -- + --------- + ---------
 |                      16       4           4    
/

$$\int x^{3} \log{\left(2 x \right)}\, dx = C + \frac{x^{4} \log{\left(x \right)}}{4} - \frac{x^{4}}{16} + \frac{x^{4} \log{\left(2 \right)}}{4}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

  1    log(2)
- -- + ------
  16     4

$$- \frac{1}{16} + \frac{\log{\left(2 \right)}}{4}$$

  1    log(2)
- -- + ------
  16     4

$$- \frac{1}{16} + \frac{\log{\left(2 \right)}}{4}$$

-1/16 + log(2)/4

Numerical answer [src]

0.110786795139986

0.110786795139986

The graph

Use the examples entering the upper and lower limits of integration.

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How to use it?

Identical expressions

Integral of x^3*ln(2x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3