Integral of xln2x dx

The solution

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

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

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

Detail solution

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

$\frac{x^{2} \left(2 \log{\left(x \right)} - 1 + \log{\left(4 \right)}\right)}{4}$
Add the constant of integration:

$\frac{x^{2} \left(2 \log{\left(x \right)} - 1 + \log{\left(4 \right)}\right)}{4}+ \mathrm{constant}$

The answer is:

$\frac{x^{2} \left(2 \log{\left(x \right)} - 1 + \log{\left(4 \right)}\right)}{4}+ \mathrm{constant}$

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

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

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

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

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

-1/4 + log(2)/2

Numerical answer [src]

0.0965735902799727

0.0965735902799727

The graph

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of xln2x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3