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

You have entered [src]

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

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

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

Detail solution

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

$2 x \log{\left(x \right)} - 2 x - \frac{\log{\left(x \right)}^{3}}{3}+ \mathrm{constant}$

The answer is:

$2 x \log{\left(x \right)} - 2 x - \frac{\log{\left(x \right)}^{3}}{3}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                       
 |                                                        
 | /                2\                   3                
 | |    (log(x) - x) |                log (x)             
 | |x - -------------| dx = C - 2*x - ------- + 2*x*log(x)
 | \          x      /                   3                
 |                                                        
/

$$\int \left(x - \frac{\left(- x + \log{\left(x \right)}\right)^{2}}{x}\right)\, dx = C + 2 x \log{\left(x \right)} - 2 x - \frac{\log{\left(x \right)}^{3}}{3}$$

Simplify

The answer [src]

-oo

$$-\infty$$

=

-oo

$$-\infty$$

-oo

Numerical answer [src]

-28570.3797156332

-28570.3797156332

Other calculators

How to use it?

Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3