Mister Exam

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Integral of d{x}:
Integral of e^x/x^2
Integral of (1+x)/x
Integral of coshx
Integral of a^x*e^x
Identical expressions
ln^ two (x)/x^ two
ln squared (x) divide by x squared
ln to the power of two (x) divide by x to the power of two
ln2(x)/x2
ln2x/x2
ln²(x)/x²
ln to the power of 2(x)/x to the power of 2
ln^2x/x^2
ln^2(x) divide by x^2
ln^2(x)/x^2dx

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

The solution

You have entered [src]

  1           
  /           
 |            
 |     2      
 |  log (x)   
 |  ------- dx
 |      2     
 |     x      
 |            
/             
0

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

Simplify

Detail solution

Let $u = \log{\left(x \right)}$ .

Then let $du = \frac{dx}{x}$ and substitute $du$ :

$\int u^{2} e^{- u}\, du$
1. There are multiple ways to do this integral.
  Method #1
  1. Let $u = - u$ .
    
    Then let $du = - du$ and substitute $- du$ :
    
    $\int u^{2} e^{u}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{2} e^{u}\right)\, du = - \int u^{2} e^{u}\, du$
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u^{2}$ 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.
      
      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.
      
      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: $2 e^{u}$
      
      So, the result is: $- u^{2} e^{u} + 2 u e^{u} - 2 e^{u}$
    Now substitute $u$ back in:
    
    $- u^{2} e^{- u} - 2 u e^{- u} - 2 e^{- u}$
  Method #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 $v{\left(u \right)}$ :
    1. There are multiple ways to do this integral.
      
      Method #1
      
      Let $u = - u$ .
      
      Then let $du = - du$ and substitute $- du$ :
      
      $\int e^{u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- e^{u}\right)\, du = - \int e^{u}\, du$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $- e^{u}$
      
      Now substitute $u$ back in:
      
      $- e^{- u}$
      
      Method #2
      
      Let $u = e^{- u}$ .
      
      Then let $du = - e^{- u} du$ and substitute $- du$ :
      
      $\int 1\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(-1\right)\, du = - \int 1\, du$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $- u$
      
      Now substitute $u$ back in:
      
      $- e^{- u}$
    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 $v{\left(u \right)}$ :
    1. Let $u = - u$ .
      
      Then let $du = - du$ and substitute $- du$ :
      
      $\int e^{u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- e^{u}\right)\, du = - \int e^{u}\, du$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $- e^{u}$
      
      Now substitute $u$ back in:
      
      $- e^{- u}$
    Now evaluate the sub-integral.
  3. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 e^{- u}\, du = 2 \int e^{- u}\, du$
    1. Let $u = - u$ .
      
      Then let $du = - du$ and substitute $- du$ :
      
      $\int e^{u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- e^{u}\right)\, du = - \int e^{u}\, du$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $- e^{u}$
      
      Now substitute $u$ back in:
      
      $- e^{- u}$
    So, the result is: $- 2 e^{- u}$
Now substitute $u$ back in:

$- \frac{\log{\left(x \right)}^{2}}{x} - \frac{2 \log{\left(x \right)}}{x} - \frac{2}{x}$
Now simplify:

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

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

The answer is:

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

The answer (Indefinite) [src]

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

$${{-\left(\log x\right)^2-2\,\log x-2}\over{x}}$$

Simplify

The answer [src]

oo

$${\it \%a}$$

oo

$$\infty$$

Simplify

Numerical answer [src]

2.55776753743723e+22

2.55776753743723e+22

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

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Identical expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #1

Method #2