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Integral of d{x}:
Integral of ex
Integral of x/x^2
Integral of ln(x)/x^3
Integral of sinx/x
Derivative of:
(1-log(x))/x^2
Limit of the function:
(1-log(x))/x^2
Identical expressions
(one -log(x))/x^ two
(1 minus logarithm of (x)) divide by x squared
(one minus logarithm of (x)) divide by x to the power of two
(1-log(x))/x2
1-logx/x2
(1-log(x))/x²
(1-log(x))/x to the power of 2
1-logx/x^2
(1-log(x)) divide by x^2
(1-log(x))/x^2dx
Similar expressions
(1+log(x))/x^2

Integral of (1-log(x))/x^2 dx

The solution

You have entered [src]

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

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

Integral((1 - log(x))/x^2, (x, 0, oo))

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

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

The answer is:

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

The answer (Indefinite) [src]

  /                          
 |                           
 | 1 - log(x)          log(x)
 | ---------- dx = C + ------
 |      2                x   
 |     x                     
 |                           
/

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

Simplify

The answer [src]

oo

$$\infty$$

oo

$$\infty$$

oo

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (1-log(x))/x^2 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3