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Integral of d{x}:
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Integral of x(x+1)^1/2
Identical expressions
x(loge(x))^ two
x( logarithm of e(x)) squared
x( logarithm of e(x)) to the power of two
x(loge(x))2
xlogex2
x(loge(x))²
x(loge(x)) to the power of 2
xlogex^2
x(loge(x))^2dx

Integral of x(loge(x))^2 dx

The solution

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

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

Integral(x*(log(x)/log(exp(1)))^2, (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $x \left(\frac{\log{\left(x \right)}}{\log{\left(e^{1} \right)}}\right)^{2} = x \log{\left(x \right)}^{2}$
2. Let $u = \log{\left(x \right)}$ .
  
  Then let $du = \frac{dx}{x}$ 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 $v{\left(u \right)}$ :
    1. 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 $v{\left(u \right)}$ :
    1. 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:
    
    $\int \frac{e^{2 u}}{2}\, du = \frac{\int e^{2 u}\, du}{2}$
    1. 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:
  
  $\frac{x^{2} \log{\left(x \right)}^{2}}{2} - \frac{x^{2} \log{\left(x \right)}}{2} + \frac{x^{2}}{4}$
Method #2
1. Rewrite the integrand:
  
  $x \left(\frac{\log{\left(x \right)}}{\log{\left(e^{1} \right)}}\right)^{2} = x \log{\left(x \right)}^{2}$
2. Let $u = \log{\left(x \right)}$ .
  
  Then let $du = \frac{dx}{x}$ 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 $v{\left(u \right)}$ :
    1. 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 $v{\left(u \right)}$ :
    1. 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:
    
    $\int \frac{e^{2 u}}{2}\, du = \frac{\int e^{2 u}\, du}{2}$
    1. 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:
  
  $\frac{x^{2} \log{\left(x \right)}^{2}}{2} - \frac{x^{2} \log{\left(x \right)}}{2} + \frac{x^{2}}{4}$
Now simplify:

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

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

The answer is:

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

The answer (Indefinite) [src]

  /                                                 
 |                                                  
 |            2           2    2    2       2       
 |   / log(x)\           x    x *log (x)   x *log(x)
 | x*|-------|  dx = C + -- + ---------- - ---------
 |   |   / 1\|           4        2            2    
 |   \log\e //                                      
 |                                                  
/

$$\int x \left(\frac{\log{\left(x \right)}}{\log{\left(e^{1} \right)}}\right)^{2}\, dx = C + \frac{x^{2} \log{\left(x \right)}^{2}}{2} - \frac{x^{2} \log{\left(x \right)}}{2} + \frac{x^{2}}{4}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

1/4

$$\frac{1}{4}$$

1/4

$$\frac{1}{4}$$

1/4

Numerical answer [src]

0.25

0.25

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

Other calculators

How to use it?

Identical expressions

Integral of x(loge(x))^2 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2