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Derivative of:
ln(1-2x)
Identical expressions
ln(one -2x)
ln(1 minus 2x)
ln(one minus 2x)
ln1-2x
ln(1-2x)dx
Similar expressions
ln(1+2x)
ln(1-2*x^3)/x

Integral of ln(1-2x) dx

The solution

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

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

Integral(log(1 - 2*x), (x, -oo, oo))

Detail solution

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

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

oo - pi*I

$$\infty - i \pi$$

oo - pi*I

$$\infty - i \pi$$

oo - pi*i

Numerical answer [src]

(1.20671494141454e+21 + 4.33314941654528e+19j)

(1.20671494141454e+21 + 4.33314941654528e+19j)

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

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Integral of ln(1-2x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #1

Method #2