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Integral of d{x}:
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Identical expressions
two (xlnx-x)+ two
2(xlnx minus x) plus 2
two (xlnx minus x) plus two
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2(xlnx-x)+2dx
Similar expressions
2(xlnx-x)-2
2(xlnx+x)+2

Integral of 2(xlnx-x)+2 dx

The solution

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 |  (2*(x*log(x) - x) + 2) dx
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0

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

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

Detail solution

Integrate term-by-term:
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 2 \left(x \log{\left(x \right)} - x\right)\, dx = 2 \int \left(x \log{\left(x \right)} - x\right)\, dx$
  1. Integrate term-by-term:
    1. There are multiple ways to do this integral.
      Method #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 $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{2}\, du$
      
      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.
      
      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:
      
      $\frac{x^{2} \log{\left(x \right)}}{2} - \frac{x^{2}}{4}$
      Method #2
      
      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 $\operatorname{dv}{\left(x \right)} = x$ .
      
      Then $\operatorname{du}{\left(x \right)} = \frac{1}{x}$ .
      
      To find $v{\left(x \right)}$ :
      
      The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x\, dx = \frac{x^{2}}{2}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{x}{2}\, dx = \frac{\int x\, dx}{2}$
      
      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 times a function is the constant times the integral of the function:
      
      $\int \left(- x\right)\, dx = - \int x\, dx$
      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: $- \frac{x^{2}}{2}$
    The result is: $\frac{x^{2} \log{\left(x \right)}}{2} - \frac{3 x^{2}}{4}$
  So, the result is: $x^{2} \log{\left(x \right)} - \frac{3 x^{2}}{2}$
1. The integral of a constant is the constant times the variable of integration:
  
  $\int 2\, dx = 2 x$
The result is: $x^{2} \log{\left(x \right)} - \frac{3 x^{2}}{2} + 2 x$
Now simplify:

$\frac{x \left(2 x \log{\left(x \right)} - 3 x + 4\right)}{2}$
Add the constant of integration:

$\frac{x \left(2 x \log{\left(x \right)} - 3 x + 4\right)}{2}+ \mathrm{constant}$

The answer is:

$\frac{x \left(2 x \log{\left(x \right)} - 3 x + 4\right)}{2}+ \mathrm{constant}$

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

1/2

$$\frac{1}{2}$$

1/2

$$\frac{1}{2}$$

1/2

Numerical answer [src]

0.5

0.5

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

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Identical expressions

Similar expressions

Integral of 2(xlnx-x)+2 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2