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1/x*(lnx+1)^2

Integral of 1/x*(lnx-1)^2 dx

The solution

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

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

Integral((log(x) - 1)^2/x, (x, exp(2), oo))

Detail solution

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

$\left(\frac{\log{\left(x \right)}^{2}}{3} - \log{\left(x \right)} + 1\right) \log{\left(x \right)}$
Add the constant of integration:

$\left(\frac{\log{\left(x \right)}^{2}}{3} - \log{\left(x \right)} + 1\right) \log{\left(x \right)}+ \mathrm{constant}$

The answer is:

\left(\frac{\log{\left(x \right)}^{2}}{3} - \log{\left(x \right)} + 1\right) \log{\left(x \right)}+ \mathrm{constant}

The answer (Indefinite) [src]

  /                                                 
 |                                                  
 |             2                       3            
 | (log(x) - 1)              2      log (x)         
 | ------------- dx = C - log (x) + ------- + log(x)
 |       x                             3            
 |                                                  
/

\int \frac{\left(\log{\left(x \right)} - 1\right)^{2}}{x}\, dx = C + \frac{\log{\left(x \right)}^{3}}{3} - \log{\left(x \right)}^{2} + \log{\left(x \right)}

Simplify

The graph

Plot the graph f(x)

The answer [src]

oo

\infty

oo

\infty

oo

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

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Identical expressions

Similar expressions

Integral of 1/x*(lnx-1)^2 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3