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Integral of d{x}:
Integral of x^4/(x^2+a^2)
Integral of x^4cosx
Integral of (x^2-2x+3)ln(x)
Integral of sinxcos^2xdx
Identical expressions
(x^ two -2x+ three)ln(x)
(x squared minus 2x plus 3)ln(x)
(x to the power of two minus 2x plus three)ln(x)
(x2-2x+3)ln(x)
x2-2x+3lnx
(x²-2x+3)ln(x)
(x to the power of 2-2x+3)ln(x)
x^2-2x+3lnx
(x^2-2x+3)ln(x)dx
Similar expressions
(x^2-2x-3)ln(x)
(x^2+2x+3)ln(x)

Integral of (x^2-2x+3)ln(x) dx

The solution

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  1                         
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 |  \x  - 2*x + 3/*log(x) dx
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0

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

Simplify

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

$\frac{x \left(6 x^{2} \log{\left(x \right)} - 2 x^{2} - 18 x \log{\left(x \right)} + 9 x + 54 \log{\left(x \right)} - 54\right)}{18}$
Add the constant of integration:

$\frac{x \left(6 x^{2} \log{\left(x \right)} - 2 x^{2} - 18 x \log{\left(x \right)} + 9 x + 54 \log{\left(x \right)} - 54\right)}{18}+ \mathrm{constant}$

The answer is:

$\frac{x \left(6 x^{2} \log{\left(x \right)} - 2 x^{2} - 18 x \log{\left(x \right)} + 9 x + 54 \log{\left(x \right)} - 54\right)}{18}+ \mathrm{constant}$

The answer (Indefinite) [src]

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

$$\left({{x^3}\over{3}}-x^2+3\,x\right)\,\log x-{{2\,x^3-9\,x^2+54\,x }\over{18}}$$

Simplify

The answer [src]

-47 
----
 18

$$-{{47}\over{18}}$$

-47 
----
 18

$$- \frac{47}{18}$$

Simplify

Numerical answer [src]

-2.61111111111111

-2.61111111111111

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (x^2-2x+3)ln(x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #2

Method #3

Method #4