Integral of xln3x dx

The solution

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  1              
  /              
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 |  x*log(3*x) dx
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0

$$\int\limits_{0}^{1} x \log{\left(3 x \right)}\, dx$$

Integral(x*log(3*x), (x, 0, 1))

Detail solution

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

$\frac{x^{2} \cdot \left(2 \log{\left(x \right)} - 1 + \log{\left(9 \right)}\right)}{4}$
Add the constant of integration:

$\frac{x^{2} \cdot \left(2 \log{\left(x \right)} - 1 + \log{\left(9 \right)}\right)}{4}+ \mathrm{constant}$

The answer is:

$\frac{x^{2} \cdot \left(2 \log{\left(x \right)} - 1 + \log{\left(9 \right)}\right)}{4}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                     2    2           2       
 |                     x    x *log(3)   x *log(x)
 | x*log(3*x) dx = C - -- + --------- + ---------
 |                     4        2           2    
/

$${{x^2\,\log \left(3\,x\right)}\over{2}}-{{x^2}\over{4}}$$

Simplify

The answer [src]

  1   log(3)
- - + ------
  4     2

$${{18\,\log 3-9}\over{36}}$$

  1   log(3)
- - + ------
  4     2

$$- \frac{1}{4} + \frac{\log{\left(3 \right)}}{2}$$

Simplify

Numerical answer [src]

0.299306144334055

0.299306144334055

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

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Identical expressions

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Integral of xln3x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #2

Method #3