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x*ln((one -x)/(one +x))
x multiply by ln((1 minus x) divide by (1 plus x))
x multiply by ln((one minus x) divide by (one plus x))
xln((1-x)/(1+x))
xln1-x/1+x
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x*ln((1-x)/(1+x))dx
Similar expressions
x*ln((1+x)/(1+x))
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xln((1-x)/(1+x))

Solving integrals/ x*ln((1-x)/(1+x))

Integral of x*ln((1-x)/(1+x)) dx

The solution

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  1                
  /                
 |                 
 |       /1 - x\   
 |  x*log|-----| dx
 |       \1 + x/   
 |                 
/                  
0

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

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

Detail solution

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

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

-1

$$-1$$

-1

$$-1$$

-1

Numerical answer [src]

-1.0

-1.0

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 x*ln((1-x)/(1+x)) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3