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Derivative of:
xln(x)-x+1
Identical expressions
xln(x)-x+ one
xln(x) minus x plus 1
xln(x) minus x plus one
xlnx-x+1
xln(x)-x+1dx
Similar expressions
xln(x)-x-1
xln(x)+x+1

Integral of xln(x)-x+1 dx

The solution

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  1                      
  /                      
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 |  (x*log(x) - x + 1) dx
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/                        
0

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

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

Detail solution

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

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

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

The answer is:

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

The answer (Indefinite) [src]

  /                                   2    2       
 |                                 3*x    x *log(x)
 | (x*log(x) - x + 1) dx = C + x - ---- + ---------
 |                                  4         2    
/

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

1/4

$$\frac{1}{4}$$

1/4

$$\frac{1}{4}$$

1/4

Numerical answer [src]

0.25

0.25

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

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Identical expressions

Similar expressions

Integral of xln(x)-x+1 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2