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Similar expressions
4x^3+2xlnx

Integral of 4x^3-2xlnx dx

The solution

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 |  \4*x  - 2*x*log(x)/ dx
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0

$$\int\limits_{0}^{1} \left(4 x^{3} - 2 x \log{\left(x \right)}\right)\, dx$$

Integral(4*x^3 - 2*x*log(x), (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:
  
  $\int 4 x^{3}\, dx = 4 \int x^{3}\, dx$
  1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
    
    $\int x^{3}\, dx = \frac{x^{4}}{4}$
  So, the result is: $x^{4}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 2 x \log{\left(x \right)}\right)\, dx = - \int 2 x \log{\left(x \right)}\, dx$
  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. 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 $v{\left(u \right)}$ :
      
      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: $\frac{e^{2 u}}{4}$
      
      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: $\frac{x^{2}}{4}$
    So, the result is: $x^{2} \log{\left(x \right)} - \frac{x^{2}}{2}$
  So, the result is: $- x^{2} \log{\left(x \right)} + \frac{x^{2}}{2}$
The result is: $x^{4} - x^{2} \log{\left(x \right)} + \frac{x^{2}}{2}$
Now simplify:

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

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

The answer is:

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

The answer (Indefinite) [src]

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

$$x^4-2\,\left({{x^2\,\log x}\over{2}}-{{x^2}\over{4}}\right)$$

Simplify

The answer [src]

3/2

$${{3}\over{2}}$$

3/2

$$\frac{3}{2}$$

Упростить

Numerical answer [src]

1.5

1.5

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

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Identical expressions

Similar expressions

Integral of 4x^3-2xlnx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #2