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(log(x)+ one)^ three /x
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(log(x)+1)3/x
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(log(x)+1)^3/xdx
Similar expressions
(log(x)-1)^3/x

Integral of (log(x)+1)^3/x dx

The solution

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  E                 
  /                 
 |                  
 |              3   
 |  (log(x) + 1)    
 |  ------------- dx
 |        x         
 |                  
/                   
1

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

Integral((log(x) + 1)^3/x, (x, 1, E))

Detail solution

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

$\frac{\left(\log{\left(x \right)} + 1\right)^{4}}{4}$
Add the constant of integration:

$\frac{\left(\log{\left(x \right)} + 1\right)^{4}}{4}+ \mathrm{constant}$

The answer is:

$\frac{\left(\log{\left(x \right)} + 1\right)^{4}}{4}+ \mathrm{constant}$

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

15/4

$$\frac{15}{4}$$

15/4

$$\frac{15}{4}$$

15/4

Numerical answer [src]

3.75

3.75

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 (log(x)+1)^3/x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3