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Integral of d{x}:
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Integral of x*dx/(x^2+1)
Identical expressions
(x- two)*log2(x)
(x minus 2) multiply by logarithm of 2(x)
(x minus two) multiply by logarithm of 2(x)
(x-2)log2(x)
x-2log2x
(x-2)*log2(x)dx
Similar expressions
(x+2)*log2(x)

Integral of (x-2)*log2(x) dx

The solution

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

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

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

Detail solution

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

$\frac{x \left(2 x \log{\left(x \right)} - x - 8 \log{\left(x \right)} + 8\right)}{4 \log{\left(2 \right)}}$
Add the constant of integration:

$\frac{x \left(2 x \log{\left(x \right)} - x - 8 \log{\left(x \right)} + 8\right)}{4 \log{\left(2 \right)}}+ \mathrm{constant}$

The answer is:

$\frac{x \left(2 x \log{\left(x \right)} - x - 8 \log{\left(x \right)} + 8\right)}{4 \log{\left(2 \right)}}+ \mathrm{constant}$

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

   7    
--------
4*log(2)

$$\frac{7}{4 \log{\left(2 \right)}}$$

   7    
--------
4*log(2)

$$\frac{7}{4 \log{\left(2 \right)}}$$

7/(4*log(2))

Numerical answer [src]

2.52471632155569

2.52471632155569

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

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Identical expressions

Similar expressions

Integral of (x-2)*log2(x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3