Mister Exam

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Identical expressions
y*(lny- one)^ two
y multiply by (lny minus 1) squared
y multiply by (lny minus one) to the power of two
y*(lny-1)2
y*lny-12
y*(lny-1)²
y*(lny-1) to the power of 2
y(lny-1)^2
y(lny-1)2
ylny-12
ylny-1^2
Similar expressions
y*(lny+1)^2

Integral of y*(lny-1)^2 dy

The solution

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  1                   
  /                   
 |                    
 |                2   
 |  y*(log(y) - 1)  dy
 |                    
/                     
0

$$\int\limits_{0}^{1} y \left(\log{\left(y \right)} - 1\right)^{2}\, dy$$

Integral(y*(log(y) - 1)^2, (y, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $y \left(\log{\left(y \right)} - 1\right)^{2} = y \log{\left(y \right)}^{2} - 2 y \log{\left(y \right)} + y$
2. Integrate term-by-term:
  1. Let $u = \log{\left(y \right)}$ .
    
    Then let $du = \frac{dy}{y}$ and substitute $du$ :
    
    $\int u^{2} e^{2 u}\, du$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u^{2}$ 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 $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{2}\, du$
      
      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.
    2. 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.
    3. 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:
    
    $\frac{y^{2} \log{\left(y \right)}^{2}}{2} - \frac{y^{2} \log{\left(y \right)}}{2} + \frac{y^{2}}{4}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 2 y \log{\left(y \right)}\right)\, dy = - 2 \int y \log{\left(y \right)}\, dy$
    1. Let $u = \log{\left(y \right)}$ .
      
      Then let $du = \frac{dy}{y}$ 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:
      
      $\frac{y^{2} \log{\left(y \right)}}{2} - \frac{y^{2}}{4}$
    So, the result is: $- y^{2} \log{\left(y \right)} + \frac{y^{2}}{2}$
  1. The integral of $y^{n}$ is $\frac{y^{n + 1}}{n + 1}$ when $n \neq -1$ :
    
    $\int y\, dy = \frac{y^{2}}{2}$
  The result is: $\frac{y^{2} \log{\left(y \right)}^{2}}{2} - \frac{3 y^{2} \log{\left(y \right)}}{2} + \frac{5 y^{2}}{4}$
Method #2
1. Rewrite the integrand:
  
  $y \left(\log{\left(y \right)} - 1\right)^{2} = y \log{\left(y \right)}^{2} - 2 y \log{\left(y \right)} + y$
2. Integrate term-by-term:
  1. Let $u = \log{\left(y \right)}$ .
    
    Then let $du = \frac{dy}{y}$ and substitute $du$ :
    
    $\int u^{2} e^{2 u}\, du$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u^{2}$ 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 $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{2}\, du$
      
      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.
    2. 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.
    3. 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:
    
    $\frac{y^{2} \log{\left(y \right)}^{2}}{2} - \frac{y^{2} \log{\left(y \right)}}{2} + \frac{y^{2}}{4}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 2 y \log{\left(y \right)}\right)\, dy = - 2 \int y \log{\left(y \right)}\, dy$
    1. Let $u = \log{\left(y \right)}$ .
      
      Then let $du = \frac{dy}{y}$ 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:
      
      $\frac{y^{2} \log{\left(y \right)}}{2} - \frac{y^{2}}{4}$
    So, the result is: $- y^{2} \log{\left(y \right)} + \frac{y^{2}}{2}$
  1. The integral of $y^{n}$ is $\frac{y^{n + 1}}{n + 1}$ when $n \neq -1$ :
    
    $\int y\, dy = \frac{y^{2}}{2}$
  The result is: $\frac{y^{2} \log{\left(y \right)}^{2}}{2} - \frac{3 y^{2} \log{\left(y \right)}}{2} + \frac{5 y^{2}}{4}$
Now simplify:

$\frac{y^{2} \left(2 \log{\left(y \right)}^{2} - 6 \log{\left(y \right)} + 5\right)}{4}$
Add the constant of integration:

$\frac{y^{2} \left(2 \log{\left(y \right)}^{2} - 6 \log{\left(y \right)} + 5\right)}{4}+ \mathrm{constant}$

The answer is:

$\frac{y^{2} \left(2 \log{\left(y \right)}^{2} - 6 \log{\left(y \right)} + 5\right)}{4}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                        
 |                             2    2    2         2       
 |               2          5*y    y *log (y)   3*y *log(y)
 | y*(log(y) - 1)  dy = C + ---- + ---------- - -----------
 |                           4         2             2     
/

$$\int y \left(\log{\left(y \right)} - 1\right)^{2}\, dy = C + \frac{y^{2} \log{\left(y \right)}^{2}}{2} - \frac{3 y^{2} \log{\left(y \right)}}{2} + \frac{5 y^{2}}{4}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

5/4

$$\frac{5}{4}$$

5/4

$$\frac{5}{4}$$

5/4

Numerical answer [src]

1.25

1.25

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

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Identical expressions

Similar expressions

Integral of y*(lny-1)^2 dy

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2