Integral of (x+2)lnx dx

The solution

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

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

$\frac{x \left(2 x \log{\left(x \right)} - x + 8 \log{\left(x \right)} - 8\right)}{4}$
Add the constant of integration:

$\frac{x \left(2 x \log{\left(x \right)} - x + 8 \log{\left(x \right)} - 8\right)}{4}+ \mathrm{constant}$

The answer is:

$\frac{x \left(2 x \log{\left(x \right)} - x + 8 \log{\left(x \right)} - 8\right)}{4}+ \mathrm{constant}$

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

-9/4

$$- \frac{9}{4}$$

-9/4

$$- \frac{9}{4}$$

-9/4

Numerical answer [src]

-2.25

-2.25

The graph

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (x+2)lnx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Method #4