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(ln^ four)(3x+ one)/(3x+ one)
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(ln^4)(3x+1)/(3x+1)dx
Similar expressions
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Integral of (ln^4)(3x+1)/(3x+1) dx

The solution

You have entered [src]

  1                     
  /                     
 |                      
 |     4                
 |  log (x)*(3*x + 1)   
 |  ----------------- dx
 |       3*x + 1        
 |                      
/                       
0

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

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

Detail solution

Let $u = \log{\left(x \right)}$ .

Then let $du = \frac{dx}{x}$ and substitute $du$ :

$\int u^{4} e^{u}\, du$
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(u \right)} = u^{4}$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$ .
  
  Then $\operatorname{du}{\left(u \right)} = 4 u^{3}$ .
  
  To find $v{\left(u \right)}$ :
  1. The integral of the exponential function is itself.
    
    $\int e^{u}\, du = e^{u}$
  Now evaluate the sub-integral.
2. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(u \right)} = 4 u^{3}$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$ .
  
  Then $\operatorname{du}{\left(u \right)} = 12 u^{2}$ .
  
  To find $v{\left(u \right)}$ :
  1. The integral of the exponential function is itself.
    
    $\int e^{u}\, du = e^{u}$
  Now evaluate the sub-integral.
3. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(u \right)} = 12 u^{2}$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$ .
  
  Then $\operatorname{du}{\left(u \right)} = 24 u$ .
  
  To find $v{\left(u \right)}$ :
  1. The integral of the exponential function is itself.
    
    $\int e^{u}\, du = e^{u}$
  Now evaluate the sub-integral.
4. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(u \right)} = 24 u$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$ .
  
  Then $\operatorname{du}{\left(u \right)} = 24$ .
  
  To find $v{\left(u \right)}$ :
  1. The integral of the exponential function is itself.
    
    $\int e^{u}\, du = e^{u}$
  Now evaluate the sub-integral.
5. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 24 e^{u}\, du = 24 \int e^{u}\, du$
  1. The integral of the exponential function is itself.
    
    $\int e^{u}\, du = e^{u}$
  So, the result is: $24 e^{u}$
Now substitute $u$ back in:

$x \log{\left(x \right)}^{4} - 4 x \log{\left(x \right)}^{3} + 12 x \log{\left(x \right)}^{2} - 24 x \log{\left(x \right)} + 24 x$
Now simplify:

$x \left(\log{\left(x \right)}^{4} - 4 \log{\left(x \right)}^{3} + 12 \log{\left(x \right)}^{2} - 24 \log{\left(x \right)} + 24\right)$
Add the constant of integration:

$x \left(\log{\left(x \right)}^{4} - 4 \log{\left(x \right)}^{3} + 12 \log{\left(x \right)}^{2} - 24 \log{\left(x \right)} + 24\right)+ \mathrm{constant}$

The answer is:

$x \left(\log{\left(x \right)}^{4} - 4 \log{\left(x \right)}^{3} + 12 \log{\left(x \right)}^{2} - 24 \log{\left(x \right)} + 24\right)+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                                      
 |                                                                                       
 |    4                                                                                  
 | log (x)*(3*x + 1)                      4                           3              2   
 | ----------------- dx = C + 24*x + x*log (x) - 24*x*log(x) - 4*x*log (x) + 12*x*log (x)
 |      3*x + 1                                                                          
 |                                                                                       
/

$$\int \frac{\left(3 x + 1\right) \log{\left(x \right)}^{4}}{3 x + 1}\, dx = C + x \log{\left(x \right)}^{4} - 4 x \log{\left(x \right)}^{3} + 12 x \log{\left(x \right)}^{2} - 24 x \log{\left(x \right)} + 24 x$$

Simplify

The answer [src]

$$24$$

Упростить

Numerical answer [src]

23.9999999999997

23.9999999999997

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

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Integral of (ln^4)(3x+1)/(3x+1) dx

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Piecewise:

The solution