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Identical expressions
((x^ five -2x)*ln3x)/x
((x to the power of 5 minus 2x) multiply by ln3x) divide by x
((x to the power of five minus 2x) multiply by ln3x) divide by x
((x5-2x)*ln3x)/x
x5-2x*ln3x/x
((x⁵-2x)*ln3x)/x
((x^5-2x)ln3x)/x
((x5-2x)ln3x)/x
x5-2xln3x/x
x^5-2xln3x/x
((x^5-2x)*ln3x) divide by x
((x^5-2x)*ln3x)/xdx
Similar expressions
((x^5+2x)*ln3x)/x

Integral of ((x^5-2x)*ln3x)/x dx

The solution

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  1                       
  /                       
 |                        
 |  / 5      \            
 |  \x  - 2*x/*log(3*x)   
 |  ------------------- dx
 |           x            
 |                        
/                         
3

$$\int\limits_{3}^{1} \frac{\left(x^{5} - 2 x\right) \log{\left(3 x \right)}}{x}\, dx$$

Integral(((x^5 - 2*x)*log(3*x))/x, (x, 3, 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 $du$ :
  
  $\int \frac{2 u^{4} \log{\left(\frac{1}{u} \right)} + 2 u^{4} \log{\left(3 \right)} - \log{\left(\frac{1}{u} \right)} - \log{\left(3 \right)}}{u^{6}}\, du$
  1. Let $u = \log{\left(\frac{1}{u} \right)}$ .
    
    Then let $du = - \frac{du}{u}$ and substitute $du$ :
    
    $\int \left(u e^{5 u} - 2 u e^{u} + e^{5 u} \log{\left(3 \right)} - 2 e^{u} \log{\left(3 \right)}\right)\, du$
    1. Integrate term-by-term:
      
      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 $\operatorname{dv}{\left(u \right)} = e^{5 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = 5 u$ .
      
      Then let $du = 5 du$ and substitute $\frac{du}{5}$ :
      
      $\int \frac{e^{u}}{5}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{5}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{5 u}}{5}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{5 u}}{5}\, du = \frac{\int e^{5 u}\, du}{5}$
      
      Let $u = 5 u$ .
      
      Then let $du = 5 du$ and substitute $\frac{du}{5}$ :
      
      $\int \frac{e^{u}}{5}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{5}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{5 u}}{5}$
      
      So, the result is: $\frac{e^{5 u}}{25}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 u e^{u}\right)\, du = - 2 \int u e^{u}\, du$
      
      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.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $- 2 u e^{u} + 2 e^{u}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int e^{5 u} \log{\left(3 \right)}\, du = \log{\left(3 \right)} \int e^{5 u}\, du$
      
      Let $u = 5 u$ .
      
      Then let $du = 5 du$ and substitute $\frac{du}{5}$ :
      
      $\int \frac{e^{u}}{5}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{5}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{5 u}}{5}$
      
      So, the result is: $\frac{e^{5 u} \log{\left(3 \right)}}{5}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 e^{u} \log{\left(3 \right)}\right)\, du = - 2 \log{\left(3 \right)} \int e^{u}\, du$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $- 2 e^{u} \log{\left(3 \right)}$
      
      The result is: $\frac{u e^{5 u}}{5} - 2 u e^{u} - \frac{e^{5 u}}{25} + \frac{e^{5 u} \log{\left(3 \right)}}{5} - 2 e^{u} \log{\left(3 \right)} + 2 e^{u}$
    Now substitute $u$ back in:
    
    $- \frac{2 \log{\left(\frac{1}{u} \right)}}{u} - \frac{2 \log{\left(3 \right)}}{u} + \frac{2}{u} + \frac{\log{\left(\frac{1}{u} \right)}}{5 u^{5}} - \frac{1}{25 u^{5}} + \frac{\log{\left(3 \right)}}{5 u^{5}}$
  Now substitute $u$ back in:
  
  $\frac{x^{5} \log{\left(x \right)}}{5} - \frac{x^{5}}{25} + \frac{x^{5} \log{\left(3 \right)}}{5} - 2 x \log{\left(x \right)} - 2 x \log{\left(3 \right)} + 2 x$
Method #2
1. Rewrite the integrand:
  
  $\frac{\left(x^{5} - 2 x\right) \log{\left(3 x \right)}}{x} = x^{4} \log{\left(x \right)} + x^{4} \log{\left(3 \right)} - 2 \log{\left(x \right)} - 2 \log{\left(3 \right)}$
2. Integrate term-by-term:
  1. Let $u = \log{\left(x \right)}$ .
    
    Then let $du = \frac{dx}{x}$ and substitute $du$ :
    
    $\int u e^{5 u}\, du$
    1. 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 $\operatorname{dv}{\left(u \right)} = e^{5 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = 5 u$ .
      
      Then let $du = 5 du$ and substitute $\frac{du}{5}$ :
      
      $\int \frac{e^{u}}{5}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{5}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{5 u}}{5}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{5 u}}{5}\, du = \frac{\int e^{5 u}\, du}{5}$
      
      Let $u = 5 u$ .
      
      Then let $du = 5 du$ and substitute $\frac{du}{5}$ :
      
      $\int \frac{e^{u}}{5}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{5}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{5 u}}{5}$
      
      So, the result is: $\frac{e^{5 u}}{25}$
    Now substitute $u$ back in:
    
    $\frac{x^{5} \log{\left(x \right)}}{5} - \frac{x^{5}}{25}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int x^{4} \log{\left(3 \right)}\, dx = \log{\left(3 \right)} \int x^{4}\, dx$
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x^{4}\, dx = \frac{x^{5}}{5}$
    So, the result is: $\frac{x^{5} \log{\left(3 \right)}}{5}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 2 \log{\left(x \right)}\right)\, dx = - 2 \int \log{\left(x \right)}\, dx$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = \log{\left(x \right)}$ and let $\operatorname{dv}{\left(x \right)} = 1$ .
      
      Then $\operatorname{du}{\left(x \right)} = \frac{1}{x}$ .
      
      To find $v{\left(x \right)}$ :
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, dx = x$
      
      Now evaluate the sub-integral.
    2. The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, dx = x$
    So, the result is: $- 2 x \log{\left(x \right)} + 2 x$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \left(- 2 \log{\left(3 \right)}\right)\, dx = - 2 x \log{\left(3 \right)}$
  The result is: $\frac{x^{5} \log{\left(x \right)}}{5} - \frac{x^{5}}{25} + \frac{x^{5} \log{\left(3 \right)}}{5} - 2 x \log{\left(x \right)} - 2 x \log{\left(3 \right)} + 2 x$
Method #3
1. Rewrite the integrand:
  
  $\frac{\left(x^{5} - 2 x\right) \log{\left(3 x \right)}}{x} = x^{4} \log{\left(x \right)} + x^{4} \log{\left(3 \right)} - 2 \log{\left(x \right)} - 2 \log{\left(3 \right)}$
2. Integrate term-by-term:
  1. Let $u = \log{\left(x \right)}$ .
    
    Then let $du = \frac{dx}{x}$ and substitute $du$ :
    
    $\int u e^{5 u}\, du$
    1. 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 $\operatorname{dv}{\left(u \right)} = e^{5 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = 5 u$ .
      
      Then let $du = 5 du$ and substitute $\frac{du}{5}$ :
      
      $\int \frac{e^{u}}{5}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{5}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{5 u}}{5}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{5 u}}{5}\, du = \frac{\int e^{5 u}\, du}{5}$
      
      Let $u = 5 u$ .
      
      Then let $du = 5 du$ and substitute $\frac{du}{5}$ :
      
      $\int \frac{e^{u}}{5}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{5}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{5 u}}{5}$
      
      So, the result is: $\frac{e^{5 u}}{25}$
    Now substitute $u$ back in:
    
    $\frac{x^{5} \log{\left(x \right)}}{5} - \frac{x^{5}}{25}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int x^{4} \log{\left(3 \right)}\, dx = \log{\left(3 \right)} \int x^{4}\, dx$
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x^{4}\, dx = \frac{x^{5}}{5}$
    So, the result is: $\frac{x^{5} \log{\left(3 \right)}}{5}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 2 \log{\left(x \right)}\right)\, dx = - 2 \int \log{\left(x \right)}\, dx$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = \log{\left(x \right)}$ and let $\operatorname{dv}{\left(x \right)} = 1$ .
      
      Then $\operatorname{du}{\left(x \right)} = \frac{1}{x}$ .
      
      To find $v{\left(x \right)}$ :
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, dx = x$
      
      Now evaluate the sub-integral.
    2. The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, dx = x$
    So, the result is: $- 2 x \log{\left(x \right)} + 2 x$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \left(- 2 \log{\left(3 \right)}\right)\, dx = - 2 x \log{\left(3 \right)}$
  The result is: $\frac{x^{5} \log{\left(x \right)}}{5} - \frac{x^{5}}{25} + \frac{x^{5} \log{\left(3 \right)}}{5} - 2 x \log{\left(x \right)} - 2 x \log{\left(3 \right)} + 2 x$
Now simplify:

$\frac{x \left(5 x^{4} \log{\left(x \right)} - x^{4} + x^{4} \log{\left(243 \right)} - 50 \log{\left(x \right)} - \log{\left(717897987691852588770249 \right)} + 50\right)}{25}$
Add the constant of integration:

$\frac{x \left(5 x^{4} \log{\left(x \right)} - x^{4} + x^{4} \log{\left(243 \right)} - 50 \log{\left(x \right)} - \log{\left(717897987691852588770249 \right)} + 50\right)}{25}+ \mathrm{constant}$

The answer is:

$\frac{x \left(5 x^{4} \log{\left(x \right)} - x^{4} + x^{4} \log{\left(243 \right)} - 50 \log{\left(x \right)} - \log{\left(717897987691852588770249 \right)} + 50\right)}{25}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                                       
 |                                                                                        
 | / 5      \                          5                              5           5       
 | \x  - 2*x/*log(3*x)                x                              x *log(3)   x *log(x)
 | ------------------- dx = C + 2*x - -- - 2*x*log(3) - 2*x*log(x) + --------- + ---------
 |          x                         25                                 5           5    
 |                                                                                        
/

$$\int \frac{\left(x^{5} - 2 x\right) \log{\left(3 x \right)}}{x}\, dx = C + \frac{x^{5} \log{\left(x \right)}}{5} - \frac{x^{5}}{25} + \frac{x^{5} \log{\left(3 \right)}}{5} - 2 x \log{\left(x \right)} - 2 x \log{\left(3 \right)} + 2 x$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

142   213*log(9)   9*log(3)
--- - ---------- - --------
 25       5           5

$$- \frac{213 \log{\left(9 \right)}}{5} - \frac{9 \log{\left(3 \right)}}{5} + \frac{142}{25}$$

142   213*log(9)   9*log(3)
--- - ---------- - --------
 25       5           5

$$- \frac{213 \log{\left(9 \right)}}{5} - \frac{9 \log{\left(3 \right)}}{5} + \frac{142}{25}$$

142/25 - 213*log(9)/5 - 9*log(3)/5

Numerical answer [src]

-89.8992691141255

-89.8992691141255

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

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How to use it?

Identical expressions

Similar expressions

Integral of ((x^5-2x)*ln3x)/x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3