Mister Exam

Lang:

Other calculators

How to use it?
Integral of d{x}:
Integral of x^(-1/2)
Integral of sin²xcos²x
Integral of x*2^(-x)
Integral of y^(-2/3)
Identical expressions
(ln^3x+ one)*(one /x)
(ln cubed x plus 1) multiply by (1 divide by x)
(ln cubed x plus one) multiply by (one divide by x)
(ln3x+1)*(1/x)
ln3x+1*1/x
(ln³x+1)*(1/x)
(ln to the power of 3x+1)*(1/x)
(ln^3x+1)(1/x)
(ln3x+1)(1/x)
ln3x+11/x
ln^3x+11/x
(ln^3x+1)*(1 divide by x)
(ln^3x+1)*(1/x)dx
Similar expressions
(ln^3x-1)*(1/x)

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

The solution

You have entered [src]

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

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

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

Detail solution

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The answer [src]

-oo

$$-\infty$$

-oo

$$-\infty$$

-oo

Numerical answer [src]

-944592.477815508

-944592.477815508

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

Other calculators

How to use it?

Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2