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Identical expressions
ln(one +x)-lnx
ln(1 plus x) minus lnx
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ln(1+x)-lnxdx
Similar expressions
ln(1+x)+lnx
ln(1-x)-lnx

Integral of ln(1+x)-lnx dx

The solution

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  4                         
  /                         
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 |  (log(1 + x) - log(x)) dx
 |                          
/                           
1

$$\int\limits_{1}^{4} \left(- \log{\left(x \right)} + \log{\left(x + 1 \right)}\right)\, dx$$

Integral(log(1 + x) - log(x), (x, 1, 4))

Detail solution

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

log(10)   9*log(4)   3*log(2)   9*log(5)
------- - -------- - -------- + --------
   2         2          2          2

$$- \frac{9 \log{\left(4 \right)}}{2} - \frac{3 \log{\left(2 \right)}}{2} + \frac{\log{\left(10 \right)}}{2} + \frac{9 \log{\left(5 \right)}}{2}$$

log(10)   9*log(4)   3*log(2)   9*log(5)
------- - -------- - -------- + --------
   2         2          2          2

$$- \frac{9 \log{\left(4 \right)}}{2} - \frac{3 \log{\left(2 \right)}}{2} + \frac{\log{\left(10 \right)}}{2} + \frac{9 \log{\left(5 \right)}}{2}$$

log(10)/2 - 9*log(4)/2 - 3*log(2)/2 + 9*log(5)/2

Numerical answer [src]

1.11571775657105

1.11571775657105

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

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Identical expressions

Similar expressions

Integral of ln(1+x)-lnx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2