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log^ two (x)/2x
logarithm of squared (x) divide by 2x
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log2(x)/2x
log2x/2x
log²(x)/2x
log to the power of 2(x)/2x
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log^2(x) divide by 2x
log^2(x)/2xdx

Solving integrals/ log^2(x)/2x

Integral of log^2(x)/2x dx

The solution

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

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

Integral(log(x)^2*x/2, (x, 1, E))

Detail solution

The integral of a constant times a function is the constant times the integral of the function:

$\int \frac{x \log{\left(x \right)}^{2}}{2}\, dx = \frac{\int x \log{\left(x \right)}^{2}\, dx}{2}$
1. 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. There are multiple ways to do this integral.
      Method #1
      
      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:
      
      $\frac{e^{2 u}}{2}$
      Method #2
      
      Let $u = e^{2 u}$ .
      
      Then let $du = 2 e^{2 u} du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{1}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{2}\, du = \frac{\int 1\, du}{2}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $\frac{u}{2}$
      
      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 $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{4}\, du$
      1. 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.
  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 $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{4}\, du$
      1. 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: $\frac{x^{2} \log{\left(x \right)}^{2}}{4} - \frac{x^{2} \log{\left(x \right)}}{4} + \frac{x^{2}}{8}$
Now simplify:

$\frac{x^{2} \cdot \left(2 \log{\left(x \right)}^{2} - 2 \log{\left(x \right)} + 1\right)}{8}$
Add the constant of integration:

$\frac{x^{2} \cdot \left(2 \log{\left(x \right)}^{2} - 2 \log{\left(x \right)} + 1\right)}{8}+ \mathrm{constant}$

The answer is:

\frac{x^{2} \cdot \left(2 \log{\left(x \right)}^{2} - 2 \log{\left(x \right)} + 1\right)}{8}+ \mathrm{constant}

The answer (Indefinite) [src]

  /                                              
 |                                               
 |    2                2    2           2    2   
 | log (x)*x          x    x *log(x)   x *log (x)
 | --------- dx = C + -- - --------- + ----------
 |     2              8        4           4     
 |                                               
/

\int \frac{x \log{\left(x \right)}^{2}}{2}\, dx = C + \frac{x^{2} \log{\left(x \right)}^{2}}{4} - \frac{x^{2} \log{\left(x \right)}}{4} + \frac{x^{2}}{8}

Simplify

The graph

Plot the graph f(x)

The answer [src]

- \frac{1}{8} + \frac{e^{2}}{8}

- \frac{1}{8} + \frac{e^{2}}{8}

Упростить

Numerical answer [src]

0.798632012366331

0.798632012366331

The graph

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

Other calculators

How to use it?

Identical expressions

Integral of log^2(x)/2x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2