Integral of (x^2-3x)lnxdx dx

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 |  \x  - 3*x/*log(x) dx
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0

\int\limits_{0}^{1} \left(x^{2} - 3 x\right) \log{\left(x \right)}\, dx

Integral((x^2 - 3*x)*log(x), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = \log{\left(x \right)}$ .
  
  Then let $du = \frac{dx}{x}$ and substitute $du$ :
  
  $\int \left(u e^{3 u} - 3 u e^{2 u}\right)\, du$
  1. Integrate term-by-term:
    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 $\operatorname{dv}{\left(u \right)} = e^{3 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = 3 u$ .
      
      Then let $du = 3 du$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{3}\, du$
      
      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: $\frac{e^{u}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{3 u}}{3}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{3 u}}{3}\, du = \frac{\int e^{3 u}\, du}{3}$
      
      Let $u = 3 u$ .
      
      Then let $du = 3 du$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{3}\, du$
      
      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: $\frac{e^{u}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{3 u}}{3}$
      
      So, the result is: $\frac{e^{3 u}}{9}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 3 u e^{2 u}\right)\, du = - 3 \int u e^{2 u}\, du$
      
      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 $\operatorname{dv}{\left(u \right)} = e^{2 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{2}\, du$
      
      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: $\frac{e^{u}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{2 u}}{2}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{2 u}}{2}\, du = \frac{\int e^{2 u}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{2}\, du$
      
      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: $\frac{e^{u}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{2 u}}{2}$
      
      So, the result is: $\frac{e^{2 u}}{4}$
      
      So, the result is: $- \frac{3 u e^{2 u}}{2} + \frac{3 e^{2 u}}{4}$
    The result is: $\frac{u e^{3 u}}{3} - \frac{3 u e^{2 u}}{2} - \frac{e^{3 u}}{9} + \frac{3 e^{2 u}}{4}$
  Now substitute $u$ back in:
  
  $\frac{x^{3} \log{\left(x \right)}}{3} - \frac{x^{3}}{9} - \frac{3 x^{2} \log{\left(x \right)}}{2} + \frac{3 x^{2}}{4}$
Method #2
1. Rewrite the integrand:
  
  $\left(x^{2} - 3 x\right) \log{\left(x \right)} = x^{2} \log{\left(x \right)} - 3 x \log{\left(x \right)}$
2. Integrate term-by-term:
  1. Let $u = \log{\left(x \right)}$ .
    
    Then let $du = \frac{dx}{x}$ and substitute $du$ :
    
    $\int u e^{3 u}\, du$
    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 $\operatorname{dv}{\left(u \right)} = e^{3 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = 3 u$ .
      
      Then let $du = 3 du$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{3}\, du$
      
      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: $\frac{e^{u}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{3 u}}{3}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{3 u}}{3}\, du = \frac{\int e^{3 u}\, du}{3}$
      
      Let $u = 3 u$ .
      
      Then let $du = 3 du$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{3}\, du$
      
      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: $\frac{e^{u}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{3 u}}{3}$
      
      So, the result is: $\frac{e^{3 u}}{9}$
    Now substitute $u$ back in:
    
    $\frac{x^{3} \log{\left(x \right)}}{3} - \frac{x^{3}}{9}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 3 x \log{\left(x \right)}\right)\, dx = - 3 \int x \log{\left(x \right)}\, dx$
    1. Let $u = \log{\left(x \right)}$ .
      
      Then let $du = \frac{dx}{x}$ and substitute $du$ :
      
      $\int u e^{2 u}\, du$
      
      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 $\operatorname{dv}{\left(u \right)} = e^{2 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{2}\, du$
      
      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: $\frac{e^{u}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{2 u}}{2}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{2 u}}{2}\, du = \frac{\int e^{2 u}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{2}\, du$
      
      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: $\frac{e^{u}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{2 u}}{2}$
      
      So, the result is: $\frac{e^{2 u}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{x^{2} \log{\left(x \right)}}{2} - \frac{x^{2}}{4}$
    So, the result is: $- \frac{3 x^{2} \log{\left(x \right)}}{2} + \frac{3 x^{2}}{4}$
  The result is: $\frac{x^{3} \log{\left(x \right)}}{3} - \frac{x^{3}}{9} - \frac{3 x^{2} \log{\left(x \right)}}{2} + \frac{3 x^{2}}{4}$
Method #3
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 $\operatorname{dv}{\left(x \right)} = x^{2} - 3 x$ .
  
  Then $\operatorname{du}{\left(x \right)} = \frac{1}{x}$ .
  
  To find $v{\left(x \right)}$ :
  1. Rewrite the integrand:
    
    $x \left(x - 3\right) = x^{2} - 3 x$
  2. Integrate term-by-term:
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x^{2}\, dx = \frac{x^{3}}{3}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 3 x\right)\, dx = - 3 \int x\, dx$
      
      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: $- \frac{3 x^{2}}{2}$
    The result is: $\frac{x^{3}}{3} - \frac{3 x^{2}}{2}$
  Now evaluate the sub-integral.
2. Rewrite the integrand:
  
  $\frac{\frac{x^{3}}{3} - \frac{3 x^{2}}{2}}{x} = \frac{x^{2}}{3} - \frac{3 x}{2}$
3. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{x^{2}}{3}\, dx = \frac{\int x^{2}\, dx}{3}$
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x^{2}\, dx = \frac{x^{3}}{3}$
    So, the result is: $\frac{x^{3}}{9}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{3 x}{2}\right)\, dx = - \frac{3 \int x\, dx}{2}$
    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: $- \frac{3 x^{2}}{4}$
  The result is: $\frac{x^{3}}{9} - \frac{3 x^{2}}{4}$
Method #4
1. Rewrite the integrand:
  
  $\left(x^{2} - 3 x\right) \log{\left(x \right)} = x^{2} \log{\left(x \right)} - 3 x \log{\left(x \right)}$
2. Integrate term-by-term:
  1. Let $u = \log{\left(x \right)}$ .
    
    Then let $du = \frac{dx}{x}$ and substitute $du$ :
    
    $\int u e^{3 u}\, du$
    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 $\operatorname{dv}{\left(u \right)} = e^{3 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = 3 u$ .
      
      Then let $du = 3 du$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{3}\, du$
      
      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: $\frac{e^{u}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{3 u}}{3}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{3 u}}{3}\, du = \frac{\int e^{3 u}\, du}{3}$
      
      Let $u = 3 u$ .
      
      Then let $du = 3 du$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{3}\, du$
      
      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: $\frac{e^{u}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{3 u}}{3}$
      
      So, the result is: $\frac{e^{3 u}}{9}$
    Now substitute $u$ back in:
    
    $\frac{x^{3} \log{\left(x \right)}}{3} - \frac{x^{3}}{9}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 3 x \log{\left(x \right)}\right)\, dx = - 3 \int x \log{\left(x \right)}\, dx$
    1. Let $u = \log{\left(x \right)}$ .
      
      Then let $du = \frac{dx}{x}$ and substitute $du$ :
      
      $\int u e^{2 u}\, du$
      
      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 $\operatorname{dv}{\left(u \right)} = e^{2 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{2}\, du$
      
      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: $\frac{e^{u}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{2 u}}{2}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{2 u}}{2}\, du = \frac{\int e^{2 u}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{2}\, du$
      
      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: $\frac{e^{u}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{2 u}}{2}$
      
      So, the result is: $\frac{e^{2 u}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{x^{2} \log{\left(x \right)}}{2} - \frac{x^{2}}{4}$
    So, the result is: $- \frac{3 x^{2} \log{\left(x \right)}}{2} + \frac{3 x^{2}}{4}$
  The result is: $\frac{x^{3} \log{\left(x \right)}}{3} - \frac{x^{3}}{9} - \frac{3 x^{2} \log{\left(x \right)}}{2} + \frac{3 x^{2}}{4}$
Now simplify:

$\frac{x^{2} \left(12 x \log{\left(x \right)} - 4 x - 54 \log{\left(x \right)} + 27\right)}{36}$
Add the constant of integration:

$\frac{x^{2} \left(12 x \log{\left(x \right)} - 4 x - 54 \log{\left(x \right)} + 27\right)}{36}+ \mathrm{constant}$

The answer is:

\frac{x^{2} \left(12 x \log{\left(x \right)} - 4 x - 54 \log{\left(x \right)} + 27\right)}{36}+ \mathrm{constant}

The answer (Indefinite) [src]

  /                                                              
 |                             3      2      2           3       
 | / 2      \                 x    3*x    3*x *log(x)   x *log(x)
 | \x  - 3*x/*log(x) dx = C - -- + ---- - ----------- + ---------
 |                            9     4          2            3    
/

\int \left(x^{2} - 3 x\right) \log{\left(x \right)}\, dx = C + \frac{x^{3} \log{\left(x \right)}}{3} - \frac{x^{3}}{9} - \frac{3 x^{2} \log{\left(x \right)}}{2} + \frac{3 x^{2}}{4}

Simplify

The graph

Plot the graph f(x)

The answer [src]

23
--
36

\frac{23}{36}

=

23
--
36

\frac{23}{36}

23/36

Numerical answer [src]

0.638888888888889

0.638888888888889

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Integral of (x^2-3x)lnxdx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Method #4