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Integral of d{x}:
Integral of x^(-1/2)
Integral of 1-7*x^2
Integral of y^(-2/3)
Integral of xe^(1-x)
Identical expressions
(x+ one)^ two *ln^ two (x)
(x plus 1) squared multiply by ln squared (x)
(x plus one) to the power of two multiply by ln to the power of two (x)
(x+1)2*ln2(x)
x+12*ln2x
(x+1)²*ln²(x)
(x+1) to the power of 2*ln to the power of 2(x)
(x+1)^2ln^2(x)
(x+1)2ln2(x)
x+12ln2x
x+1^2ln^2x
(x+1)^2*ln^2(x)dx
Similar expressions
(x-1)^2*ln^2(x)

Integral of (x+1)^2*ln^2(x) dx

The solution

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0

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

Integral((x + 1)^2*log(x)^2, (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 $du$ :
  
  $\int \left(u^{2} e^{3 u} + 2 u^{2} e^{2 u} + u^{2} e^{u}\right)\, du$
  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 $u{\left(u \right)} = u^{2}$ and let $\operatorname{dv}{\left(u \right)} = e^{3 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 2 u$ .
      
      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. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = \frac{2 u}{3}$ and let $\operatorname{dv}{\left(u \right)} = e^{3 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = \frac{2}{3}$ .
      
      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.
    3. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{2 e^{3 u}}{9}\, du = \frac{2 \int e^{3 u}\, du}{9}$
      
      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: $\frac{2 e^{3 u}}{27}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 u^{2} e^{2 u}\, du = 2 \int u^{2} e^{2 u}\, du$
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u^{2}$ and let $\operatorname{dv}{\left(u \right)} = e^{2 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 2 u$ .
      
      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.
      
      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: $u^{2} e^{2 u} - u e^{2 u} + \frac{e^{2 u}}{2}$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u^{2}$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 2 u$ .
      
      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.
    2. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = 2 u$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 2$ .
      
      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.
    3. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 e^{u}\, du = 2 \int e^{u}\, du$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $2 e^{u}$
    The result is: $\frac{u^{2} e^{3 u}}{3} + u^{2} e^{2 u} + u^{2} e^{u} - \frac{2 u e^{3 u}}{9} - u e^{2 u} - 2 u e^{u} + \frac{2 e^{3 u}}{27} + \frac{e^{2 u}}{2} + 2 e^{u}$
  Now substitute $u$ back in:
  
  $\frac{x^{3} \log{\left(x \right)}^{2}}{3} - \frac{2 x^{3} \log{\left(x \right)}}{9} + \frac{2 x^{3}}{27} + x^{2} \log{\left(x \right)}^{2} - x^{2} \log{\left(x \right)} + \frac{x^{2}}{2} + x \log{\left(x \right)}^{2} - 2 x \log{\left(x \right)} + 2 x$
Method #2
1. Rewrite the integrand:
  
  $\left(x + 1\right)^{2} \log{\left(x \right)}^{2} = x^{2} \log{\left(x \right)}^{2} + 2 x \log{\left(x \right)}^{2} + \log{\left(x \right)}^{2}$
2. Integrate term-by-term:
  1. Let $u = \log{\left(x \right)}$ .
    
    Then let $du = \frac{dx}{x}$ and substitute $du$ :
    
    $\int u^{2} e^{3 u}\, du$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u^{2}$ and let $\operatorname{dv}{\left(u \right)} = e^{3 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 2 u$ .
      
      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. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = \frac{2 u}{3}$ and let $\operatorname{dv}{\left(u \right)} = e^{3 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = \frac{2}{3}$ .
      
      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.
    3. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{2 e^{3 u}}{9}\, du = \frac{2 \int e^{3 u}\, du}{9}$
      
      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: $\frac{2 e^{3 u}}{27}$
    Now substitute $u$ back in:
    
    $\frac{x^{3} \log{\left(x \right)}^{2}}{3} - \frac{2 x^{3} \log{\left(x \right)}}{9} + \frac{2 x^{3}}{27}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 x \log{\left(x \right)}^{2}\, dx = 2 \int x \log{\left(x \right)}^{2}\, dx$
    1. Let $u = \log{\left(x \right)}$ .
      
      Then let $du = \frac{dx}{x}$ and substitute $du$ :
      
      $\int u^{2} e^{2 u}\, du$
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u^{2}$ and let $\operatorname{dv}{\left(u \right)} = e^{2 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 2 u$ .
      
      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.
      
      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:
      
      $\frac{x^{2} \log{\left(x \right)}^{2}}{2} - \frac{x^{2} \log{\left(x \right)}}{2} + \frac{x^{2}}{4}$
    So, the result is: $x^{2} \log{\left(x \right)}^{2} - x^{2} \log{\left(x \right)} + \frac{x^{2}}{2}$
  1. Let $u = \log{\left(x \right)}$ .
    
    Then let $du = \frac{dx}{x}$ and substitute $du$ :
    
    $\int u^{2} e^{u}\, du$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u^{2}$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 2 u$ .
      
      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.
    2. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = 2 u$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 2$ .
      
      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.
    3. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 e^{u}\, du = 2 \int e^{u}\, du$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $2 e^{u}$
    Now substitute $u$ back in:
    
    $x \log{\left(x \right)}^{2} - 2 x \log{\left(x \right)} + 2 x$
  The result is: $\frac{x^{3} \log{\left(x \right)}^{2}}{3} - \frac{2 x^{3} \log{\left(x \right)}}{9} + \frac{2 x^{3}}{27} + x^{2} \log{\left(x \right)}^{2} - x^{2} \log{\left(x \right)} + \frac{x^{2}}{2} + x \log{\left(x \right)}^{2} - 2 x \log{\left(x \right)} + 2 x$
Method #3
1. Rewrite the integrand:
  
  $\left(x + 1\right)^{2} \log{\left(x \right)}^{2} = x^{2} \log{\left(x \right)}^{2} + 2 x \log{\left(x \right)}^{2} + \log{\left(x \right)}^{2}$
2. Integrate term-by-term:
  1. Let $u = \log{\left(x \right)}$ .
    
    Then let $du = \frac{dx}{x}$ and substitute $du$ :
    
    $\int u^{2} e^{3 u}\, du$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u^{2}$ and let $\operatorname{dv}{\left(u \right)} = e^{3 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 2 u$ .
      
      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. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = \frac{2 u}{3}$ and let $\operatorname{dv}{\left(u \right)} = e^{3 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = \frac{2}{3}$ .
      
      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.
    3. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{2 e^{3 u}}{9}\, du = \frac{2 \int e^{3 u}\, du}{9}$
      
      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: $\frac{2 e^{3 u}}{27}$
    Now substitute $u$ back in:
    
    $\frac{x^{3} \log{\left(x \right)}^{2}}{3} - \frac{2 x^{3} \log{\left(x \right)}}{9} + \frac{2 x^{3}}{27}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 x \log{\left(x \right)}^{2}\, dx = 2 \int x \log{\left(x \right)}^{2}\, dx$
    1. Let $u = \log{\left(x \right)}$ .
      
      Then let $du = \frac{dx}{x}$ and substitute $du$ :
      
      $\int u^{2} e^{2 u}\, du$
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u^{2}$ and let $\operatorname{dv}{\left(u \right)} = e^{2 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 2 u$ .
      
      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.
      
      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:
      
      $\frac{x^{2} \log{\left(x \right)}^{2}}{2} - \frac{x^{2} \log{\left(x \right)}}{2} + \frac{x^{2}}{4}$
    So, the result is: $x^{2} \log{\left(x \right)}^{2} - x^{2} \log{\left(x \right)} + \frac{x^{2}}{2}$
  1. Let $u = \log{\left(x \right)}$ .
    
    Then let $du = \frac{dx}{x}$ and substitute $du$ :
    
    $\int u^{2} e^{u}\, du$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u^{2}$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 2 u$ .
      
      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.
    2. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = 2 u$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 2$ .
      
      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.
    3. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 e^{u}\, du = 2 \int e^{u}\, du$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $2 e^{u}$
    Now substitute $u$ back in:
    
    $x \log{\left(x \right)}^{2} - 2 x \log{\left(x \right)} + 2 x$
  The result is: $\frac{x^{3} \log{\left(x \right)}^{2}}{3} - \frac{2 x^{3} \log{\left(x \right)}}{9} + \frac{2 x^{3}}{27} + x^{2} \log{\left(x \right)}^{2} - x^{2} \log{\left(x \right)} + \frac{x^{2}}{2} + x \log{\left(x \right)}^{2} - 2 x \log{\left(x \right)} + 2 x$
Now simplify:

$\frac{x \left(18 x^{2} \log{\left(x \right)}^{2} - 12 x^{2} \log{\left(x \right)} + 4 x^{2} + 54 x \log{\left(x \right)}^{2} - 54 x \log{\left(x \right)} + 27 x + 54 \log{\left(x \right)}^{2} - 108 \log{\left(x \right)} + 108\right)}{54}$
Add the constant of integration:

$\frac{x \left(18 x^{2} \log{\left(x \right)}^{2} - 12 x^{2} \log{\left(x \right)} + 4 x^{2} + 54 x \log{\left(x \right)}^{2} - 54 x \log{\left(x \right)} + 27 x + 54 \log{\left(x \right)}^{2} - 108 \log{\left(x \right)} + 108\right)}{54}+ \mathrm{constant}$

The answer is:

$\frac{x \left(18 x^{2} \log{\left(x \right)}^{2} - 12 x^{2} \log{\left(x \right)} + 4 x^{2} + 54 x \log{\left(x \right)}^{2} - 54 x \log{\left(x \right)} + 27 x + 54 \log{\left(x \right)}^{2} - 108 \log{\left(x \right)} + 108\right)}{54}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                                                                      
 |                            2            3                                                        3           3    2   
 |        2    2             x          2*x         2       2    2       2                       2*x *log(x)   x *log (x)
 | (x + 1) *log (x) dx = C + -- + 2*x + ---- + x*log (x) + x *log (x) - x *log(x) - 2*x*log(x) - ----------- + ----------
 |                           2           27                                                           9            3     
/

$${{x^3\,\left(9\,\left(\log x\right)^2-6\,\log x+2\right)}\over{27}} +{{x^2\,\left(2\,\left(\log x\right)^2-2\,\log x+1\right)}\over{2}}+ x\,\left(\left(\log x\right)^2-2\,\log x+2\right)$$

Simplify

The answer [src]

139
---
 54

$${{139}\over{54}}$$

139
---
 54

$$\frac{139}{54}$$

Упростить

Numerical answer [src]

2.57407407407407

2.57407407407407

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

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Identical expressions

Similar expressions

Integral of (x+1)^2*ln^2(x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #2

Method #3