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Identical expressions
(one +3x)*log4x
(1 plus 3x) multiply by logarithm of 4x
(one plus 3x) multiply by logarithm of 4x
(1+3x)log4x
1+3xlog4x
(1+3x)*log4xdx
Similar expressions
(1-3x)*log4x

Integral of (1+3x)*log4x dx

The solution

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  4                      
  /                      
 |                       
 |  (1 + 3*x)*log(4*x) dx
 |                       
/                        
1

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

Integral((1 + 3*x)*log(4*x), (x, 1, 4))

Detail solution

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

$\frac{x \left(6 x \log{\left(x \right)} - 3 x + x \log{\left(4096 \right)} + 4 \log{\left(x \right)} - 4 + \log{\left(256 \right)}\right)}{4}$
Add the constant of integration:

$\frac{x \left(6 x \log{\left(x \right)} - 3 x + x \log{\left(4096 \right)} + 4 \log{\left(x \right)} - 4 + \log{\left(256 \right)}\right)}{4}+ \mathrm{constant}$

The answer is:

$\frac{x \left(6 x \log{\left(x \right)} - 3 x + x \log{\left(4096 \right)} + 4 \log{\left(x \right)} - 4 + \log{\left(256 \right)}\right)}{4}+ \mathrm{constant}$

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

  57                5*log(4)
- -- + 28*log(16) - --------
  4                    2

$$- \frac{57}{4} - \frac{5 \log{\left(4 \right)}}{2} + 28 \log{\left(16 \right)}$$

  57                5*log(4)
- -- + 28*log(16) - --------
  4                    2

$$- \frac{57}{4} - \frac{5 \log{\left(4 \right)}}{2} + 28 \log{\left(16 \right)}$$

-57/4 + 28*log(16) - 5*log(4)/2

Numerical answer [src]

59.9167483199141

59.9167483199141

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

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How to use it?

Identical expressions

Similar expressions

Integral of (1+3x)*log4x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3