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(log(one /x))(one -x)
( logarithm of (1 divide by x))(1 minus x)
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Similar expressions
(log(1/x))(1+x)

Integral of (log(1/x))(1-x) dx

The solution

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  p                  
  -                  
  n                  
  /                  
 |                   
 |     /1\           
 |  log|-|*(1 - x) dx
 |     \x/           
 |                   
/                    
0

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

Integral(log(1/x)*(1 - x), (x, 0, p/n))

Detail solution

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

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The answer [src]

    /      2 \            2 
p   |p    p  |    /n\    p  
- + |- - ----|*log|-| - ----
n   |n      2|    \p/      2
    \    2*n /          4*n

$$\left(\frac{p}{n} - \frac{p^{2}}{2 n^{2}}\right) \log{\left(\frac{n}{p} \right)} + \frac{p}{n} - \frac{p^{2}}{4 n^{2}}$$

    /      2 \            2 
p   |p    p  |    /n\    p  
- + |- - ----|*log|-| - ----
n   |n      2|    \p/      2
    \    2*n /          4*n

$$\left(\frac{p}{n} - \frac{p^{2}}{2 n^{2}}\right) \log{\left(\frac{n}{p} \right)} + \frac{p}{n} - \frac{p^{2}}{4 n^{2}}$$

p/n + (p/n - p^2/(2*n^2))*log(n/p) - p^2/(4*n^2)

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (log(1/x))(1-x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #1

Method #2

Method #3

Method #4