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Identical expressions
(x^n- one)/log(x)
(x to the power of n minus 1) divide by logarithm of (x)
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(xn-1)/log(x)
xn-1/logx
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(x^n-1) divide by log(x)
(x^n-1)/log(x)dx
Similar expressions
(x^n+1)/log(x)

Integral of (x^n-1)/log(x) dx

The solution

You have entered [src]

  n          
  /          
 |           
 |   n       
 |  x  - 1   
 |  ------ dx
 |  log(x)   
 |           
/            
0

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

Integral((x^n - 1*1)/log(x), (x, 0, n))

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 $du$ :
  
  $\int \frac{e^{u} e^{n u} - e^{u}}{u}\, du$
  1. Let $u = \frac{1}{u}$ .
    
    Then let $du = - \frac{du}{u^{2}}$ and substitute $du$ :
    
    $\int \frac{- e^{\frac{1}{u}} e^{\frac{n}{u}} + e^{\frac{1}{u}}}{u}\, du$
    1. Rewrite the integrand:
      
      $\frac{- e^{\frac{1}{u}} e^{\frac{n}{u}} + e^{\frac{1}{u}}}{u} = - \frac{e^{\frac{1}{u}} e^{\frac{n}{u}}}{u} + \frac{e^{\frac{1}{u}}}{u}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{e^{\frac{1}{u}} e^{\frac{n}{u}}}{u}\right)\, du = - \int \frac{e^{\frac{1}{u}} e^{\frac{n}{u}}}{u}\, du$
      
      Let $u = \frac{1}{u}$ .
      
      Then let $du = - \frac{du}{u^{2}}$ and substitute $- du$ :
      
      $\int \frac{e^{u} e^{n u}}{u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{e^{u} e^{u n}}{u}\right)\, du = - \int \frac{e^{u} e^{u n}}{u}\, du$
      
      EiRule(a=n + 1, b=0, context=exp(_u)*exp(_u*n)/_u, symbol=_u)
      
      So, the result is: $- \operatorname{Ei}{\left(u \left(n + 1\right) \right)}$
      
      Now substitute $u$ back in:
      
      $- \operatorname{Ei}{\left(\frac{n + 1}{u} \right)}$
      
      So, the result is: $\operatorname{Ei}{\left(\frac{n + 1}{u} \right)}$
      
      Let $u = \frac{1}{u}$ .
      
      Then let $du = - \frac{du}{u^{2}}$ and substitute $- du$ :
      
      $\int \frac{e^{u}}{u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{e^{u}}{u}\right)\, du = - \int \frac{e^{u}}{u}\, du$
      
      EiRule(a=1, b=0, context=exp(_u)/_u, symbol=_u)
      
      So, the result is: $- \operatorname{Ei}{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- \operatorname{Ei}{\left(\frac{1}{u} \right)}$
      
      The result is: $- \operatorname{Ei}{\left(\frac{1}{u} \right)} + \operatorname{Ei}{\left(\frac{n + 1}{u} \right)}$
    Now substitute $u$ back in:
    
    $- \operatorname{Ei}{\left(u \right)} + \operatorname{Ei}{\left(u \left(n + 1\right) \right)}$
  Now substitute $u$ back in:
  
  $\operatorname{Ei}{\left(\left(n + 1\right) \log{\left(x \right)} \right)} - \operatorname{Ei}{\left(\log{\left(x \right)} \right)}$
Method #2
1. Rewrite the integrand:
  
  $\frac{x^{n} - 1}{\log{\left(x \right)}} = \frac{x^{n}}{\log{\left(x \right)}} - \frac{1}{\log{\left(x \right)}}$
2. Integrate term-by-term:
  1. Let $u = \log{\left(x \right)}$ .
    
    Then let $du = \frac{dx}{x}$ and substitute $du$ :
    
    $\int \frac{e^{u} e^{n u}}{u}\, du$
    Now substitute $u$ back in:
    
    $\operatorname{Ei}{\left(\left(n + 1\right) \log{\left(x \right)} \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{\log{\left(x \right)}}\right)\, dx = - \int \frac{1}{\log{\left(x \right)}}\, dx$
    So, the result is: $- \operatorname{li}{\left(x \right)}$
  The result is: $\operatorname{Ei}{\left(\left(n + 1\right) \log{\left(x \right)} \right)} - \operatorname{li}{\left(x \right)}$
Add the constant of integration:

$\operatorname{Ei}{\left(\left(n + 1\right) \log{\left(x \right)} \right)} - \operatorname{Ei}{\left(\log{\left(x \right)} \right)}+ \mathrm{constant}$

The answer is:

$\operatorname{Ei}{\left(\left(n + 1\right) \log{\left(x \right)} \right)} - \operatorname{Ei}{\left(\log{\left(x \right)} \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                               
 |                                                
 |  n                                             
 | x  - 1                                         
 | ------ dx = C - Ei(log(x)) + Ei((1 + n)*log(x))
 | log(x)                                         
 |                                                
/

$${{\int {{{e^{n\,\log x}}\over{\left(\log x\right)^2}}}{\;dx}}\over{ n+1}}+{{x\,e^{n\,\log x}}\over{\left(n+1\right)\,\log x}}+\Gamma \left(0 , -\log x\right)$$

The answer [src]

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

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

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

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

Simplify

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (x^n-1)/log(x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2