Integral of (2x-5)lnx dx

The solution

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$$\int\limits_{0}^{1} \left(2 x - 5\right) \log{\left(x \right)}\, dx$$

Integral((2*x - 1*5)*log(x), (x, 0, 1))

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 $du$ :
  
  $\int \left(2 u e^{2 u} - 5 u e^{u}\right)\, du$
  1. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 u e^{2 u}\, du = 2 \int u e^{2 u}\, du$
      
      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 $\operatorname{dv}{\left(u \right)} = e^{2 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      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.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{2 u}}{2}\, du = \frac{\int e^{2 u}\, du}{2}$
      
      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:
      
      $\frac{e^{2 u}}{2}$
      
      So, the result is: $\frac{e^{2 u}}{4}$
      
      So, the result is: $u e^{2 u} - \frac{e^{2 u}}{2}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 5 u e^{u}\right)\, du = - 5 \int u e^{u}\, du$
      
      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: $- 5 u e^{u} + 5 e^{u}$
    The result is: $u e^{2 u} - 5 u e^{u} - \frac{e^{2 u}}{2} + 5 e^{u}$
  Now substitute $u$ back in:
  
  $x^{2} \log{\left(x \right)} - \frac{x^{2}}{2} - 5 x \log{\left(x \right)} + 5 x$
Method #2
1. Rewrite the integrand:
  
  $\left(2 x - 5\right) \log{\left(x \right)} = 2 x \log{\left(x \right)} - 5 \log{\left(x \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 x \log{\left(x \right)}\, dx = 2 \int x \log{\left(x \right)}\, dx$
    1. Let $u = \log{\left(x \right)}$ .
      
      Then let $du = \frac{dx}{x}$ and substitute $du$ :
      
      $\int u e^{2 u}\, du$
      
      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 $\operatorname{dv}{\left(u \right)} = e^{2 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      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:
      
      $\frac{e^{2 u}}{2}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{2 u}}{2}\, du = \frac{\int e^{2 u}\, du}{2}$
      
      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:
      
      $\frac{e^{2 u}}{2}$
      
      So, the result is: $\frac{e^{2 u}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{x^{2} \log{\left(x \right)}}{2} - \frac{x^{2}}{4}$
    So, the result is: $x^{2} \log{\left(x \right)} - \frac{x^{2}}{2}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 5 \log{\left(x \right)}\right)\, dx = - 5 \int \log{\left(x \right)}\, dx$
    1. 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.
    2. The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, dx = x$
    So, the result is: $- 5 x \log{\left(x \right)} + 5 x$
  The result is: $x^{2} \log{\left(x \right)} - \frac{x^{2}}{2} - 5 x \log{\left(x \right)} + 5 x$
Method #3
1. 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 $\operatorname{dv}{\left(x \right)} = 2 x - 5$ .
  
  Then $\operatorname{du}{\left(x \right)} = \frac{1}{x}$ .
  
  To find $v{\left(x \right)}$ :
  1. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 x\, dx = 2 \int x\, dx$
      
      The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x\, dx = \frac{x^{2}}{2}$
      
      So, the result is: $x^{2}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \left(-5\right)\, dx = - 5 x$
    The result is: $x^{2} - 5 x$
  Now evaluate the sub-integral.
2. Rewrite the integrand:
  
  $\frac{x^{2} - 5 x}{x} = x - 5$
3. 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 is the constant times the variable of integration:
    
    $\int \left(-5\right)\, dx = - 5 x$
  The result is: $\frac{x^{2}}{2} - 5 x$
Method #4
1. Rewrite the integrand:
  
  $\left(2 x - 5\right) \log{\left(x \right)} = 2 x \log{\left(x \right)} - 5 \log{\left(x \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 x \log{\left(x \right)}\, dx = 2 \int x \log{\left(x \right)}\, dx$
    1. Let $u = \log{\left(x \right)}$ .
      
      Then let $du = \frac{dx}{x}$ and substitute $du$ :
      
      $\int u e^{2 u}\, du$
      
      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 $\operatorname{dv}{\left(u \right)} = e^{2 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      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:
      
      $\frac{e^{2 u}}{2}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{2 u}}{2}\, du = \frac{\int e^{2 u}\, du}{2}$
      
      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:
      
      $\frac{e^{2 u}}{2}$
      
      So, the result is: $\frac{e^{2 u}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{x^{2} \log{\left(x \right)}}{2} - \frac{x^{2}}{4}$
    So, the result is: $x^{2} \log{\left(x \right)} - \frac{x^{2}}{2}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 5 \log{\left(x \right)}\right)\, dx = - 5 \int \log{\left(x \right)}\, dx$
    1. 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.
    2. The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, dx = x$
    So, the result is: $- 5 x \log{\left(x \right)} + 5 x$
  The result is: $x^{2} \log{\left(x \right)} - \frac{x^{2}}{2} - 5 x \log{\left(x \right)} + 5 x$
Now simplify:

$\frac{x \left(2 x \log{\left(x \right)} - x - 10 \log{\left(x \right)} + 10\right)}{2}$
Add the constant of integration:

$\frac{x \left(2 x \log{\left(x \right)} - x - 10 \log{\left(x \right)} + 10\right)}{2}+ \mathrm{constant}$

The answer is:

$\frac{x \left(2 x \log{\left(x \right)} - x - 10 \log{\left(x \right)} + 10\right)}{2}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                 2                         
 |                                 x     2                    
 | (2*x - 5)*log(x) dx = C + 5*x - -- + x *log(x) - 5*x*log(x)
 |                                 2                          
/

$$\left(x^2-5\,x\right)\,\log x-{{x^2-10\,x}\over{2}}$$

Simplify

The answer [src]

9/2

$${{9}\over{2}}$$

9/2

$$\frac{9}{2}$$

Упростить

Numerical answer [src]

4.5

4.5

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

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Identical expressions

Similar expressions

Integral of (2x-5)lnx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #2

Method #3

Method #4