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Identical expressions
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lnydy/y1-ln2*y
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Similar expressions
(lnydy)/(y(1+((ln^2*y))))

Solving integrals/ (lnydy)/(y(1-((ln^2*y))))

Integral of (lnydy)/(y(1-((ln^2*y)))) dy

The solution

You have entered [src]

  1                   
  /                   
 |                    
 |       log(y)       
 |  --------------- dy
 |    /       2   \   
 |  y*\1 - log (y)/   
 |                    
/                     
0

$$\int\limits_{0}^{1} \frac{\log{\left(y \right)}}{y \left(1 - \log{\left(y \right)}^{2}\right)}\, dy$$

Integral(log(y)/((y*(1 - log(y)^2))), (y, 0, 1))

Detail solution

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

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

The answer is:

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

The answer (Indefinite) [src]

  /                                          
 |                             /        2   \
 |      log(y)              log\-1 + log (y)/
 | --------------- dy = C - -----------------
 |   /       2   \                  2        
 | y*\1 - log (y)/                           
 |                                           
/

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

nan

$$\text{NaN}$$

nan

$$\text{NaN}$$

nan

Numerical answer [src]

22.919249968258

22.919249968258

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

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Identical expressions

Similar expressions

Integral of (lnydy)/(y(1-((ln^2*y)))) dy

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #1

Method #2

Method #3