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Identical expressions
dx/x*(one +ln(x))^ five
dx divide by x multiply by (1 plus ln(x)) to the power of 5
dx divide by x multiply by (one plus ln(x)) to the power of five
dx/x*(1+ln(x))5
dx/x*1+lnx5
dx/x*(1+ln(x))⁵
dx/x(1+ln(x))^5
dx/x(1+ln(x))5
dx/x1+lnx5
dx/x1+lnx^5
dx divide by x*(1+ln(x))^5
Similar expressions
dx/x*(1-ln(x))^5

Integral of dx/x*(1+ln(x))^5 dx

The solution

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  1                 
  /                 
 |                  
 |              5   
 |  (1 + log(x))    
 |  ------------- dx
 |        x         
 |                  
/                   
0

$$\int\limits_{0}^{1} \frac{\left(\log{\left(x \right)} + 1\right)^{5}}{x}\, dx$$

Integral((1 + log(x))^5/x, (x, 0, 1))

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 $- du$ :
  
  $\int \left(- \frac{\log{\left(\frac{1}{u} \right)}^{5} + 5 \log{\left(\frac{1}{u} \right)}^{4} + 10 \log{\left(\frac{1}{u} \right)}^{3} + 10 \log{\left(\frac{1}{u} \right)}^{2} + 5 \log{\left(\frac{1}{u} \right)} + 1}{u}\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\log{\left(\frac{1}{u} \right)}^{5} + 5 \log{\left(\frac{1}{u} \right)}^{4} + 10 \log{\left(\frac{1}{u} \right)}^{3} + 10 \log{\left(\frac{1}{u} \right)}^{2} + 5 \log{\left(\frac{1}{u} \right)} + 1}{u}\, du = - \int \frac{\log{\left(\frac{1}{u} \right)}^{5} + 5 \log{\left(\frac{1}{u} \right)}^{4} + 10 \log{\left(\frac{1}{u} \right)}^{3} + 10 \log{\left(\frac{1}{u} \right)}^{2} + 5 \log{\left(\frac{1}{u} \right)} + 1}{u}\, du$
    1. Let $u = \frac{1}{u}$ .
      
      Then let $du = - \frac{du}{u^{2}}$ and substitute $- du$ :
      
      $\int \left(- \frac{\log{\left(u \right)}^{5} + 5 \log{\left(u \right)}^{4} + 10 \log{\left(u \right)}^{3} + 10 \log{\left(u \right)}^{2} + 5 \log{\left(u \right)} + 1}{u}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\log{\left(u \right)}^{5} + 5 \log{\left(u \right)}^{4} + 10 \log{\left(u \right)}^{3} + 10 \log{\left(u \right)}^{2} + 5 \log{\left(u \right)} + 1}{u}\, du = - \int \frac{\log{\left(u \right)}^{5} + 5 \log{\left(u \right)}^{4} + 10 \log{\left(u \right)}^{3} + 10 \log{\left(u \right)}^{2} + 5 \log{\left(u \right)} + 1}{u}\, du$
      
      Let $u = \log{\left(u \right)}$ .
      
      Then let $du = \frac{du}{u}$ and substitute $du$ :
      
      $\int \left(u^{5} + 5 u^{4} + 10 u^{3} + 10 u^{2} + 5 u + 1\right)\, du$
      
      Integrate term-by-term:
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{5}\, du = \frac{u^{6}}{6}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 5 u^{4}\, du = 5 \int u^{4}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{4}\, du = \frac{u^{5}}{5}$
      
      So, the result is: $u^{5}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 10 u^{3}\, du = 10 \int u^{3}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{3}\, du = \frac{u^{4}}{4}$
      
      So, the result is: $\frac{5 u^{4}}{2}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 10 u^{2}\, du = 10 \int u^{2}\, du$
      
      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: $\frac{10 u^{3}}{3}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 5 u\, du = 5 \int u\, du$
      
      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: $\frac{5 u^{2}}{2}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      The result is: $\frac{u^{6}}{6} + u^{5} + \frac{5 u^{4}}{2} + \frac{10 u^{3}}{3} + \frac{5 u^{2}}{2} + u$
      
      Now substitute $u$ back in:
      
      $\frac{\log{\left(u \right)}^{6}}{6} + \log{\left(u \right)}^{5} + \frac{5 \log{\left(u \right)}^{4}}{2} + \frac{10 \log{\left(u \right)}^{3}}{3} + \frac{5 \log{\left(u \right)}^{2}}{2} + \log{\left(u \right)}$
      
      So, the result is: $- \frac{\log{\left(u \right)}^{6}}{6} - \log{\left(u \right)}^{5} - \frac{5 \log{\left(u \right)}^{4}}{2} - \frac{10 \log{\left(u \right)}^{3}}{3} - \frac{5 \log{\left(u \right)}^{2}}{2} - \log{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- \frac{\log{\left(u \right)}^{6}}{6} + \log{\left(u \right)}^{5} - \frac{5 \log{\left(u \right)}^{4}}{2} + \frac{10 \log{\left(u \right)}^{3}}{3} - \frac{5 \log{\left(u \right)}^{2}}{2} + \log{\left(u \right)}$
    So, the result is: $\frac{\log{\left(u \right)}^{6}}{6} - \log{\left(u \right)}^{5} + \frac{5 \log{\left(u \right)}^{4}}{2} - \frac{10 \log{\left(u \right)}^{3}}{3} + \frac{5 \log{\left(u \right)}^{2}}{2} - \log{\left(u \right)}$
  Now substitute $u$ back in:
  
  $\frac{\log{\left(x \right)}^{6}}{6} + \log{\left(x \right)}^{5} + \frac{5 \log{\left(x \right)}^{4}}{2} + \frac{10 \log{\left(x \right)}^{3}}{3} + \frac{5 \log{\left(x \right)}^{2}}{2} + \log{\left(x \right)}$
Method #2
1. Rewrite the integrand:
  
  $\frac{\left(\log{\left(x \right)} + 1\right)^{5}}{x} = \frac{\log{\left(x \right)}^{5} + 5 \log{\left(x \right)}^{4} + 10 \log{\left(x \right)}^{3} + 10 \log{\left(x \right)}^{2} + 5 \log{\left(x \right)} + 1}{x}$
2. Let $u = \frac{1}{x}$ .
  
  Then let $du = - \frac{dx}{x^{2}}$ and substitute $- du$ :
  
  $\int \left(- \frac{\log{\left(\frac{1}{u} \right)}^{5} + 5 \log{\left(\frac{1}{u} \right)}^{4} + 10 \log{\left(\frac{1}{u} \right)}^{3} + 10 \log{\left(\frac{1}{u} \right)}^{2} + 5 \log{\left(\frac{1}{u} \right)} + 1}{u}\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\log{\left(\frac{1}{u} \right)}^{5} + 5 \log{\left(\frac{1}{u} \right)}^{4} + 10 \log{\left(\frac{1}{u} \right)}^{3} + 10 \log{\left(\frac{1}{u} \right)}^{2} + 5 \log{\left(\frac{1}{u} \right)} + 1}{u}\, du = - \int \frac{\log{\left(\frac{1}{u} \right)}^{5} + 5 \log{\left(\frac{1}{u} \right)}^{4} + 10 \log{\left(\frac{1}{u} \right)}^{3} + 10 \log{\left(\frac{1}{u} \right)}^{2} + 5 \log{\left(\frac{1}{u} \right)} + 1}{u}\, du$
    1. Let $u = \log{\left(\frac{1}{u} \right)}$ .
      
      Then let $du = - \frac{du}{u}$ and substitute $du$ :
      
      $\int \left(- u^{5} - 5 u^{4} - 10 u^{3} - 10 u^{2} - 5 u - 1\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{5}\right)\, du = - \int u^{5}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{5}\, du = \frac{u^{6}}{6}$
      
      So, the result is: $- \frac{u^{6}}{6}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 5 u^{4}\right)\, du = - 5 \int u^{4}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{4}\, du = \frac{u^{5}}{5}$
      
      So, the result is: $- u^{5}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 10 u^{3}\right)\, du = - 10 \int u^{3}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{3}\, du = \frac{u^{4}}{4}$
      
      So, the result is: $- \frac{5 u^{4}}{2}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 10 u^{2}\right)\, du = - 10 \int u^{2}\, du$
      
      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: $- \frac{10 u^{3}}{3}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 5 u\right)\, du = - 5 \int u\, du$
      
      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: $- \frac{5 u^{2}}{2}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \left(-1\right)\, du = - u$
      
      The result is: $- \frac{u^{6}}{6} - u^{5} - \frac{5 u^{4}}{2} - \frac{10 u^{3}}{3} - \frac{5 u^{2}}{2} - u$
      
      Now substitute $u$ back in:
      
      $- \frac{\log{\left(\frac{1}{u} \right)}^{6}}{6} - \log{\left(\frac{1}{u} \right)}^{5} - \frac{5 \log{\left(\frac{1}{u} \right)}^{4}}{2} - \frac{10 \log{\left(\frac{1}{u} \right)}^{3}}{3} - \frac{5 \log{\left(\frac{1}{u} \right)}^{2}}{2} - \log{\left(\frac{1}{u} \right)}$
    So, the result is: $\frac{\log{\left(\frac{1}{u} \right)}^{6}}{6} + \log{\left(\frac{1}{u} \right)}^{5} + \frac{5 \log{\left(\frac{1}{u} \right)}^{4}}{2} + \frac{10 \log{\left(\frac{1}{u} \right)}^{3}}{3} + \frac{5 \log{\left(\frac{1}{u} \right)}^{2}}{2} + \log{\left(\frac{1}{u} \right)}$
  Now substitute $u$ back in:
  
  $\frac{\log{\left(x \right)}^{6}}{6} + \log{\left(x \right)}^{5} + \frac{5 \log{\left(x \right)}^{4}}{2} + \frac{10 \log{\left(x \right)}^{3}}{3} + \frac{5 \log{\left(x \right)}^{2}}{2} + \log{\left(x \right)}$
Method #3
1. Rewrite the integrand:
  
  $\frac{\left(\log{\left(x \right)} + 1\right)^{5}}{x} = \frac{\log{\left(x \right)}^{5}}{x} + \frac{5 \log{\left(x \right)}^{4}}{x} + \frac{10 \log{\left(x \right)}^{3}}{x} + \frac{10 \log{\left(x \right)}^{2}}{x} + \frac{5 \log{\left(x \right)}}{x} + \frac{1}{x}$
2. Integrate term-by-term:
  1. Let $u = \frac{1}{x}$ .
    
    Then let $du = - \frac{dx}{x^{2}}$ and substitute $- du$ :
    
    $\int \left(- \frac{\log{\left(\frac{1}{u} \right)}^{5}}{u}\right)\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\log{\left(\frac{1}{u} \right)}^{5}}{u}\, du = - \int \frac{\log{\left(\frac{1}{u} \right)}^{5}}{u}\, du$
      
      Let $u = \log{\left(\frac{1}{u} \right)}$ .
      
      Then let $du = - \frac{du}{u}$ and substitute $- du$ :
      
      $\int \left(- u^{5}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{5}\, du = - \int u^{5}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{5}\, du = \frac{u^{6}}{6}$
      
      So, the result is: $- \frac{u^{6}}{6}$
      
      Now substitute $u$ back in:
      
      $- \frac{\log{\left(\frac{1}{u} \right)}^{6}}{6}$
      
      So, the result is: $\frac{\log{\left(\frac{1}{u} \right)}^{6}}{6}$
    Now substitute $u$ back in:
    
    $\frac{\log{\left(x \right)}^{6}}{6}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{5 \log{\left(x \right)}^{4}}{x}\, dx = 5 \int \frac{\log{\left(x \right)}^{4}}{x}\, dx$
    1. Let $u = \frac{1}{x}$ .
      
      Then let $du = - \frac{dx}{x^{2}}$ and substitute $- du$ :
      
      $\int \left(- \frac{\log{\left(\frac{1}{u} \right)}^{4}}{u}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\log{\left(\frac{1}{u} \right)}^{4}}{u}\, du = - \int \frac{\log{\left(\frac{1}{u} \right)}^{4}}{u}\, du$
      
      Let $u = \log{\left(\frac{1}{u} \right)}$ .
      
      Then let $du = - \frac{du}{u}$ and substitute $- du$ :
      
      $\int \left(- u^{4}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{4}\, du = - \int u^{4}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{4}\, du = \frac{u^{5}}{5}$
      
      So, the result is: $- \frac{u^{5}}{5}$
      
      Now substitute $u$ back in:
      
      $- \frac{\log{\left(\frac{1}{u} \right)}^{5}}{5}$
      
      So, the result is: $\frac{\log{\left(\frac{1}{u} \right)}^{5}}{5}$
      
      Now substitute $u$ back in:
      
      $\frac{\log{\left(x \right)}^{5}}{5}$
    So, the result is: $\log{\left(x \right)}^{5}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{10 \log{\left(x \right)}^{3}}{x}\, dx = 10 \int \frac{\log{\left(x \right)}^{3}}{x}\, dx$
    1. Let $u = \frac{1}{x}$ .
      
      Then let $du = - \frac{dx}{x^{2}}$ and substitute $- du$ :
      
      $\int \left(- \frac{\log{\left(\frac{1}{u} \right)}^{3}}{u}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\log{\left(\frac{1}{u} \right)}^{3}}{u}\, du = - \int \frac{\log{\left(\frac{1}{u} \right)}^{3}}{u}\, du$
      
      Let $u = \log{\left(\frac{1}{u} \right)}$ .
      
      Then let $du = - \frac{du}{u}$ and substitute $- du$ :
      
      $\int \left(- u^{3}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{3}\, du = - \int u^{3}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{3}\, du = \frac{u^{4}}{4}$
      
      So, the result is: $- \frac{u^{4}}{4}$
      
      Now substitute $u$ back in:
      
      $- \frac{\log{\left(\frac{1}{u} \right)}^{4}}{4}$
      
      So, the result is: $\frac{\log{\left(\frac{1}{u} \right)}^{4}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\log{\left(x \right)}^{4}}{4}$
    So, the result is: $\frac{5 \log{\left(x \right)}^{4}}{2}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{10 \log{\left(x \right)}^{2}}{x}\, dx = 10 \int \frac{\log{\left(x \right)}^{2}}{x}\, dx$
    1. Let $u = \frac{1}{x}$ .
      
      Then let $du = - \frac{dx}{x^{2}}$ and substitute $- du$ :
      
      $\int \left(- \frac{\log{\left(\frac{1}{u} \right)}^{2}}{u}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\log{\left(\frac{1}{u} \right)}^{2}}{u}\, du = - \int \frac{\log{\left(\frac{1}{u} \right)}^{2}}{u}\, du$
      
      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:
      
      $\frac{\log{\left(x \right)}^{3}}{3}$
    So, the result is: $\frac{10 \log{\left(x \right)}^{3}}{3}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{5 \log{\left(x \right)}}{x}\, dx = 5 \int \frac{\log{\left(x \right)}}{x}\, dx$
    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: $\frac{5 \log{\left(x \right)}^{2}}{2}$
  1. The integral of $\frac{1}{x}$ is $\log{\left(x \right)}$ .
  The result is: $\frac{\log{\left(x \right)}^{6}}{6} + \log{\left(x \right)}^{5} + \frac{5 \log{\left(x \right)}^{4}}{2} + \frac{10 \log{\left(x \right)}^{3}}{3} + \frac{5 \log{\left(x \right)}^{2}}{2} + \log{\left(x \right)}$
Now simplify:

$\frac{\left(\log{\left(x \right)}^{5} + 6 \log{\left(x \right)}^{4} + 15 \log{\left(x \right)}^{3} + 20 \log{\left(x \right)}^{2} + 15 \log{\left(x \right)} + 6\right) \log{\left(x \right)}}{6}$
Add the constant of integration:

$\frac{\left(\log{\left(x \right)}^{5} + 6 \log{\left(x \right)}^{4} + 15 \log{\left(x \right)}^{3} + 20 \log{\left(x \right)}^{2} + 15 \log{\left(x \right)} + 6\right) \log{\left(x \right)}}{6}+ \mathrm{constant}$

The answer is:

$\frac{\left(\log{\left(x \right)}^{5} + 6 \log{\left(x \right)}^{4} + 15 \log{\left(x \right)}^{3} + 20 \log{\left(x \right)}^{2} + 15 \log{\left(x \right)} + 6\right) \log{\left(x \right)}}{6}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                                      
 |                                                                                       
 |             5                       6           2           4            3            
 | (1 + log(x))              5      log (x)   5*log (x)   5*log (x)   10*log (x)         
 | ------------- dx = C + log (x) + ------- + --------- + --------- + ---------- + log(x)
 |       x                             6          2           2           3              
 |                                                                                       
/

$$\int \frac{\left(\log{\left(x \right)} + 1\right)^{5}}{x}\, dx = C + \frac{\log{\left(x \right)}^{6}}{6} + \log{\left(x \right)}^{5} + \frac{5 \log{\left(x \right)}^{4}}{2} + \frac{10 \log{\left(x \right)}^{3}}{3} + \frac{5 \log{\left(x \right)}^{2}}{2} + \log{\left(x \right)}$$

Simplify

The answer [src]

-oo

$$-\infty$$

-oo

$$-\infty$$

-oo

Numerical answer [src]

-1066584803.24705

-1066584803.24705

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

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Identical expressions

Similar expressions

Integral of dx/x*(1+ln(x))^5 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3