Integral of (x+1)log2(x) dx

The solution

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

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

Integral((x + 1)*(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 $\frac{du}{\log{\left(2 \right)}}$ :
  
  $\int \frac{u e^{2 u} + u e^{u}}{\log{\left(2 \right)}}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(u e^{2 u} + u e^{u}\right)\, du = \frac{\int \left(u e^{2 u} + u e^{u}\right)\, du}{\log{\left(2 \right)}}$
    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: $\frac{e^{2 u}}{4}$
      
      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.
      
      $\int e^{u}\, du = e^{u}$
      
      The result is: $\frac{u e^{2 u}}{2} + u e^{u} - \frac{e^{2 u}}{4} - e^{u}$
    So, the result is: $\frac{\frac{u e^{2 u}}{2} + u e^{u} - \frac{e^{2 u}}{4} - e^{u}}{\log{\left(2 \right)}}$
  Now substitute $u$ back in:
  
  $\frac{\frac{x^{2} \log{\left(x \right)}}{2} - \frac{x^{2}}{4} + x \log{\left(x \right)} - x}{\log{\left(2 \right)}}$
Method #2
1. Rewrite the integrand:
  
  $\frac{\log{\left(x \right)}}{\log{\left(2 \right)}} \left(x + 1\right) = \frac{x \log{\left(x \right)} + \log{\left(x \right)}}{\log{\left(2 \right)}}$
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{x \log{\left(x \right)} + \log{\left(x \right)}}{\log{\left(2 \right)}}\, dx = \frac{\int \left(x \log{\left(x \right)} + \log{\left(x \right)}\right)\, dx}{\log{\left(2 \right)}}$
  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. 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:
      
      $\int 1\, dx = x$
    The result is: $\frac{x^{2} \log{\left(x \right)}}{2} - \frac{x^{2}}{4} + x \log{\left(x \right)} - x$
  So, the result is: $\frac{\frac{x^{2} \log{\left(x \right)}}{2} - \frac{x^{2}}{4} + x \log{\left(x \right)} - x}{\log{\left(2 \right)}}$
Method #3
1. Rewrite the integrand:
  
  $\frac{\log{\left(x \right)}}{\log{\left(2 \right)}} \left(x + 1\right) = \frac{x \log{\left(x \right)}}{\log{\left(2 \right)}} + \frac{\log{\left(x \right)}}{\log{\left(2 \right)}}$
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:
    
    $\int \frac{\log{\left(x \right)}}{\log{\left(2 \right)}}\, dx = \frac{\int \log{\left(x \right)}\, dx}{\log{\left(2 \right)}}$
    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: $\frac{x \log{\left(x \right)} - x}{\log{\left(2 \right)}}$
  The result is: $\frac{x \log{\left(x \right)} - x}{\log{\left(2 \right)}} + \frac{\frac{x^{2} \log{\left(x \right)}}{2} - \frac{x^{2}}{4}}{\log{\left(2 \right)}}$
Now simplify:

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

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

The answer is:

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

The answer (Indefinite) [src]

                                 2               2       
  /                             x               x *log(x)
 |                         -x - -- + x*log(x) + ---------
 |         log(x)               4                   2    
 | (x + 1)*------ dx = C + ------------------------------
 |         log(2)                      log(2)            
 |                                                       
/

$$\int \frac{\log{\left(x \right)}}{\log{\left(2 \right)}} \left(x + 1\right)\, dx = C + \frac{\frac{x^{2} \log{\left(x \right)}}{2} - \frac{x^{2}}{4} + x \log{\left(x \right)} - x}{\log{\left(2 \right)}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

  -5    
--------
4*log(2)

$$- \frac{5}{4 \log{\left(2 \right)}}$$

  -5    
--------
4*log(2)

$$- \frac{5}{4 \log{\left(2 \right)}}$$

-5/(4*log(2))

Numerical answer [src]

-1.8033688011112

-1.8033688011112

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

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Identical expressions

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Integral of (x+1)log2(x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3