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

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  1                
  /                
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 |  /    log(x)\   
 |  |1 + ------| dx
 |  \      x   /   
 |                 
/                  
0

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

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

Detail solution

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The answer [src]

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$$-\infty$$

=

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$$-\infty$$

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Numerical answer [src]

-970.963863415327

-970.963863415327

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2