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Identical expressions
3x^2log4x
3x squared logarithm of 4x
3x2log4x
3x²log4x
3x to the power of 2log4x
3x^2log4xdx

Integral of 3x^2log4x dx

The solution

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 |  3*x *log(4*x) dx
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0

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

Integral((3*x^2)*log(4*x), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $3 x^{2} \log{\left(4 x \right)} = 3 x^{2} \log{\left(x \right)} + 6 x^{2} \log{\left(2 \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 3 x^{2} \log{\left(x \right)}\, dx = 3 \int x^{2} \log{\left(x \right)}\, dx$
    1. Let $u = \log{\left(x \right)}$ .
      
      Then let $du = \frac{dx}{x}$ and substitute $du$ :
      
      $\int u e^{3 u}\, du$
      
      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 $\operatorname{dv}{\left(u \right)} = e^{3 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = 3 u$ .
      
      Then let $du = 3 du$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{3}\, du$
      
      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: $\frac{e^{u}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{3 u}}{3}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{3 u}}{3}\, du = \frac{\int e^{3 u}\, du}{3}$
      
      Let $u = 3 u$ .
      
      Then let $du = 3 du$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{3}\, du$
      
      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: $\frac{e^{u}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{3 u}}{3}$
      
      So, the result is: $\frac{e^{3 u}}{9}$
      
      Now substitute $u$ back in:
      
      $\frac{x^{3} \log{\left(x \right)}}{3} - \frac{x^{3}}{9}$
    So, the result is: $x^{3} \log{\left(x \right)} - \frac{x^{3}}{3}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 6 x^{2} \log{\left(2 \right)}\, dx = 6 \log{\left(2 \right)} \int x^{2}\, dx$
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x^{2}\, dx = \frac{x^{3}}{3}$
    So, the result is: $2 x^{3} \log{\left(2 \right)}$
  The result is: $x^{3} \log{\left(x \right)} - \frac{x^{3}}{3} + 2 x^{3} \log{\left(2 \right)}$
Method #2
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = \log{\left(4 x \right)}$ and let $\operatorname{dv}{\left(x \right)} = 3 x^{2}$ .
  
  Then $\operatorname{du}{\left(x \right)} = \frac{1}{x}$ .
  
  To find $v{\left(x \right)}$ :
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 3 x^{2}\, dx = 3 \int x^{2}\, dx$
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x^{2}\, dx = \frac{x^{3}}{3}$
    So, the result is: $x^{3}$
  Now evaluate the sub-integral.
2. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
  
  $\int x^{2}\, dx = \frac{x^{3}}{3}$
Method #3
1. Rewrite the integrand:
  
  $3 x^{2} \log{\left(4 x \right)} = 3 x^{2} \log{\left(x \right)} + 6 x^{2} \log{\left(2 \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 3 x^{2} \log{\left(x \right)}\, dx = 3 \int x^{2} \log{\left(x \right)}\, dx$
    1. Let $u = \log{\left(x \right)}$ .
      
      Then let $du = \frac{dx}{x}$ and substitute $du$ :
      
      $\int u e^{3 u}\, du$
      
      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 $\operatorname{dv}{\left(u \right)} = e^{3 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = 3 u$ .
      
      Then let $du = 3 du$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{3}\, du$
      
      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: $\frac{e^{u}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{3 u}}{3}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{3 u}}{3}\, du = \frac{\int e^{3 u}\, du}{3}$
      
      Let $u = 3 u$ .
      
      Then let $du = 3 du$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{3}\, du$
      
      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: $\frac{e^{u}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{3 u}}{3}$
      
      So, the result is: $\frac{e^{3 u}}{9}$
      
      Now substitute $u$ back in:
      
      $\frac{x^{3} \log{\left(x \right)}}{3} - \frac{x^{3}}{9}$
    So, the result is: $x^{3} \log{\left(x \right)} - \frac{x^{3}}{3}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 6 x^{2} \log{\left(2 \right)}\, dx = 6 \log{\left(2 \right)} \int x^{2}\, dx$
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x^{2}\, dx = \frac{x^{3}}{3}$
    So, the result is: $2 x^{3} \log{\left(2 \right)}$
  The result is: $x^{3} \log{\left(x \right)} - \frac{x^{3}}{3} + 2 x^{3} \log{\left(2 \right)}$
Now simplify:

$\frac{x^{3} \left(3 \log{\left(x \right)} - 1 + \log{\left(64 \right)}\right)}{3}$
Add the constant of integration:

$\frac{x^{3} \left(3 \log{\left(x \right)} - 1 + \log{\left(64 \right)}\right)}{3}+ \mathrm{constant}$

The answer is:

$\frac{x^{3} \left(3 \log{\left(x \right)} - 1 + \log{\left(64 \right)}\right)}{3}+ \mathrm{constant}$

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

-1/3 + log(4)

$$- \frac{1}{3} + \log{\left(4 \right)}$$

-1/3 + log(4)

$$- \frac{1}{3} + \log{\left(4 \right)}$$

-1/3 + log(4)

Numerical answer [src]

1.05296102778656

1.05296102778656

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

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How to use it?

Identical expressions

Integral of 3x^2log4x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3