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Limit of the function:
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Identical expressions
(one -log(x))/x
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(1-log(x))/xdx
Similar expressions
(1+log(x))/x

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

The solution

You have entered [src]

  x              
 e               
  /              
 |               
 |  1 - log(x)   
 |  ---------- dx
 |      x        
 |               
/                
1

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

Integral((1 - log(x))/x, (x, 1, exp(x)))

Detail solution

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

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The answer [src]

     2/ x\          
  log \e /      / x\
- -------- + log\e /
     2

$$- \frac{\log{\left(e^{x} \right)}^{2}}{2} + \log{\left(e^{x} \right)}$$

     2/ x\          
  log \e /      / x\
- -------- + log\e /
     2

$$- \frac{\log{\left(e^{x} \right)}^{2}}{2} + \log{\left(e^{x} \right)}$$

-log(exp(x))^2/2 + log(exp(x))

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

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Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3