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Identical expressions
(ln2*x)/(x*ln4*x)
(ln2 multiply by x) divide by (x multiply by ln4 multiply by x)
(ln2x)/(xln4x)
ln2x/xln4x
(ln2*x) divide by (x*ln4*x)
(ln2*x)/(x*ln4*x)dx

Integral of (ln2x)/(xln4*x) dx

The solution

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  1              
  /              
 |               
 |   log(2*x)    
 |  ---------- dx
 |  x*log(4*x)   
 |               
/                
0

$$\int\limits_{0}^{1} \frac{\log{\left(2 x \right)}}{x \log{\left(4 x \right)}}\, dx$$

Integral(log(2*x)/((x*log(4*x))), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\frac{\log{\left(2 x \right)}}{x \log{\left(4 x \right)}} = \frac{\log{\left(x \right)} + \log{\left(2 \right)}}{x \log{\left(x \right)} + 2 x \log{\left(2 \right)}}$
2. Let $u = \log{\left(x \right)}$ .
  
  Then let $du = \frac{dx}{x}$ and substitute $du$ :
  
  $\int \frac{u + \log{\left(2 \right)}}{u + 2 \log{\left(2 \right)}}\, du$
  1. Rewrite the integrand:
    
    $\frac{u + \log{\left(2 \right)}}{u + 2 \log{\left(2 \right)}} = 1 - \frac{\log{\left(2 \right)}}{u + 2 \log{\left(2 \right)}}$
  2. Integrate term-by-term:
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\log{\left(2 \right)}}{u + 2 \log{\left(2 \right)}}\right)\, du = - \log{\left(2 \right)} \int \frac{1}{u + 2 \log{\left(2 \right)}}\, du$
      
      Let $u = u + 2 \log{\left(2 \right)}$ .
      
      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:
      
      $\log{\left(u + 2 \log{\left(2 \right)} \right)}$
      
      So, the result is: $- \log{\left(2 \right)} \log{\left(u + 2 \log{\left(2 \right)} \right)}$
    The result is: $u - \log{\left(2 \right)} \log{\left(u + 2 \log{\left(2 \right)} \right)}$
  Now substitute $u$ back in:
  
  $\log{\left(x \right)} - \log{\left(2 \right)} \log{\left(\log{\left(x \right)} + 2 \log{\left(2 \right)} \right)}$
Method #2
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = \log{\left(2 x \right)}$ and let $\operatorname{dv}{\left(x \right)} = \frac{1}{x \log{\left(4 x \right)}}$ .
  
  Then $\operatorname{du}{\left(x \right)} = \frac{1}{x}$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = \log{\left(4 x \right)}$ .
    
    Then let $du = \frac{dx}{x}$ and substitute $du$ :
    
    $\int \frac{1}{u}\, du$
    1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
    Now substitute $u$ back in:
    
    $\log{\left(\log{\left(4 x \right)} \right)}$
  Now evaluate the sub-integral.
2. Let $u = \frac{1}{x}$ .
  
  Then let $du = - \frac{dx}{x^{2}}$ and substitute $- du$ :
  
  $\int \left(- \frac{\log{\left(\log{\left(\frac{1}{u} \right)} + 2 \log{\left(2 \right)} \right)}}{u}\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\log{\left(\log{\left(\frac{1}{u} \right)} + 2 \log{\left(2 \right)} \right)}}{u}\, du = - \int \frac{\log{\left(\log{\left(\frac{1}{u} \right)} + 2 \log{\left(2 \right)} \right)}}{u}\, du$
    1. Let $u = \frac{1}{u}$ .
      
      Then let $du = - \frac{du}{u^{2}}$ and substitute $- du$ :
      
      $\int \left(- \frac{\log{\left(\log{\left(u \right)} + 2 \log{\left(2 \right)} \right)}}{u}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\log{\left(\log{\left(u \right)} + 2 \log{\left(2 \right)} \right)}}{u}\, du = - \int \frac{\log{\left(\log{\left(u \right)} + 2 \log{\left(2 \right)} \right)}}{u}\, du$
      
      Let $u = \log{\left(\log{\left(u \right)} + 2 \log{\left(2 \right)} \right)}$ .
      
      Then let $du = \frac{du}{u \left(\log{\left(u \right)} + 2 \log{\left(2 \right)}\right)}$ and substitute $du$ :
      
      $\int u e^{u}\, du$
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      Now evaluate the sub-integral.
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      Now substitute $u$ back in:
      
      $\left(\log{\left(u \right)} + 2 \log{\left(2 \right)}\right) \log{\left(\log{\left(u \right)} + 2 \log{\left(2 \right)} \right)} - \log{\left(u \right)} - 2 \log{\left(2 \right)}$
      
      So, the result is: $- \left(\log{\left(u \right)} + 2 \log{\left(2 \right)}\right) \log{\left(\log{\left(u \right)} + 2 \log{\left(2 \right)} \right)} + \log{\left(u \right)} + 2 \log{\left(2 \right)}$
      
      Now substitute $u$ back in:
      
      $- \left(- \log{\left(u \right)} + 2 \log{\left(2 \right)}\right) \log{\left(- \log{\left(u \right)} + 2 \log{\left(2 \right)} \right)} - \log{\left(u \right)} + 2 \log{\left(2 \right)}$
    So, the result is: $\left(- \log{\left(u \right)} + 2 \log{\left(2 \right)}\right) \log{\left(- \log{\left(u \right)} + 2 \log{\left(2 \right)} \right)} + \log{\left(u \right)} - 2 \log{\left(2 \right)}$
  Now substitute $u$ back in:
  
  $\left(\log{\left(x \right)} + 2 \log{\left(2 \right)}\right) \log{\left(\log{\left(x \right)} + 2 \log{\left(2 \right)} \right)} - \log{\left(x \right)} - 2 \log{\left(2 \right)}$
Now simplify:

$\log{\left(x \right)} - \log{\left(2 \right)} \log{\left(\log{\left(4 x \right)} \right)}$
Add the constant of integration:

$\log{\left(x \right)} - \log{\left(2 \right)} \log{\left(\log{\left(4 x \right)} \right)}+ \mathrm{constant}$

The answer is:

$\log{\left(x \right)} - \log{\left(2 \right)} \log{\left(\log{\left(4 x \right)} \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                          
 |                                                           
 |  log(2*x)                                                 
 | ---------- dx = C - log(2)*log(2*log(2) + log(x)) + log(x)
 | x*log(4*x)                                                
 |                                                           
/

$$\int \frac{\log{\left(2 x \right)}}{x \log{\left(4 x \right)}}\, dx = C + \log{\left(x \right)} - \log{\left(2 \right)} \log{\left(\log{\left(x \right)} + 2 \log{\left(2 \right)} \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

nan

$$\text{NaN}$$

nan

$$\text{NaN}$$

nan

Numerical answer [src]

59.6786882263115

59.6786882263115

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

Other calculators

How to use it?

Identical expressions

Integral of (ln2x)/(xln4*x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Other calculators

How to use it?

Identical expressions

Integral of (ln2*x)/(x*ln4*x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Integral of (ln2x)/(xln4*x) dx