Integral of xlog(x)-x+c1x+c2 dx

The solution

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 |  (x*log(x) - x + c1*x + c2) dx
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$$\int\limits_{0}^{1} \left(c_{2} + \left(c_{1} x + \left(x \log{\left(x \right)} - x\right)\right)\right)\, dx$$

Integral(x*log(x) - x + c1*x + c2, (x, 0, 1))

Detail solution

Integrate term-by-term:
1. The integral of a constant is the constant times the variable of integration:
  
  $\int c_{2}\, dx = c_{2} x$
1. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int c_{1} x\, dx = c_{1} \int x\, dx$
    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: $\frac{c_{1} x^{2}}{2}$
  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: $\frac{x^{2} \log{\left(x \right)}}{2} - \frac{3 x^{2}}{4}$
  The result is: $\frac{c_{1} x^{2}}{2} + \frac{x^{2} \log{\left(x \right)}}{2} - \frac{3 x^{2}}{4}$
The result is: $\frac{c_{1} x^{2}}{2} + c_{2} x + \frac{x^{2} \log{\left(x \right)}}{2} - \frac{3 x^{2}}{4}$
Now simplify:

$\frac{x \left(2 c_{1} x + 4 c_{2} + 2 x \log{\left(x \right)} - 3 x\right)}{4}$
Add the constant of integration:

$\frac{x \left(2 c_{1} x + 4 c_{2} + 2 x \log{\left(x \right)} - 3 x\right)}{4}+ \mathrm{constant}$

The answer is:

$\frac{x \left(2 c_{1} x + 4 c_{2} + 2 x \log{\left(x \right)} - 3 x\right)}{4}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                       2              2    2       
 |                                     3*x           c1*x    x *log(x)
 | (x*log(x) - x + c1*x + c2) dx = C - ---- + c2*x + ----- + ---------
 |                                      4              2         2    
/

$$\int \left(c_{2} + \left(c_{1} x + \left(x \log{\left(x \right)} - x\right)\right)\right)\, dx = C + \frac{c_{1} x^{2}}{2} + c_{2} x + \frac{x^{2} \log{\left(x \right)}}{2} - \frac{3 x^{2}}{4}$$

Simplify

The answer [src]

  3        c1
- - + c2 + --
  4        2

$$\frac{c_{1}}{2} + c_{2} - \frac{3}{4}$$

  3        c1
- - + c2 + --
  4        2

$$\frac{c_{1}}{2} + c_{2} - \frac{3}{4}$$

-3/4 + c2 + c1/2

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

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Integral of xlog(x)-x+c1x+c2 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2