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Integral of d{x}:
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Integral of e^(x+2)
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Integral of cos²(x)
Identical expressions
lgx/x^ three
lgx divide by x cubed
lgx divide by x to the power of three
lgx/x3
lgx/x³
lgx/x to the power of 3
lgx divide by x^3
lgx/x^3dx

Integral of lgx/x^3 dx

The solution

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

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

Integral(log(x)/(x^3), (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 $du$ :
  
  $\int u e^{- 2 u}\, du$
  1. Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(u \right)} = u$ and let $\operatorname{dv}{\left(u \right)} = e^{- 2 u}$ .
    
    Then $\operatorname{du}{\left(u \right)} = 1$ .
    
    To find $v{\left(u \right)}$ :
    1. There are multiple ways to do this integral.
      
      Method #1
      
      Let $u = - 2 u$ .
      
      Then let $du = - 2 du$ and substitute $- \frac{du}{2}$ :
      
      $\int \frac{e^{u}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{e^{u}}{2}\right)\, du = - \frac{\int e^{u}\, du}{2}$
      
      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:
      
      $- \frac{e^{- 2 u}}{2}$
      
      Method #2
      
      Let $u = e^{- 2 u}$ .
      
      Then let $du = - 2 e^{- 2 u} du$ and substitute $- \frac{du}{2}$ :
      
      $\int \frac{1}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{2}\right)\, du = - \frac{\int 1\, du}{2}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $- \frac{u}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{e^{- 2 u}}{2}$
    Now evaluate the sub-integral.
  2. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{e^{- 2 u}}{2}\right)\, du = - \frac{\int e^{- 2 u}\, du}{2}$
    1. Let $u = - 2 u$ .
      
      Then let $du = - 2 du$ and substitute $- \frac{du}{2}$ :
      
      $\int \frac{e^{u}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{e^{u}}{2}\right)\, du = - \frac{\int e^{u}\, du}{2}$
      
      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:
  
  $- \frac{\log{\left(x \right)}}{2 x^{2}} - \frac{1}{4 x^{2}}$
Method #2
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = \log{\left(x \right)}$ and let $\operatorname{dv}{\left(x \right)} = \frac{1}{x^{3}}$ .
  
  Then $\operatorname{du}{\left(x \right)} = \frac{1}{x}$ .
  
  To find $v{\left(x \right)}$ :
  1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
    
    $\int \frac{1}{x^{3}}\, dx = - \frac{1}{2 x^{2}}$
  Now evaluate the sub-integral.
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \frac{1}{2 x^{3}}\right)\, dx = - \frac{\int \frac{1}{x^{3}}\, dx}{2}$
  1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
    
    $\int \frac{1}{x^{3}}\, dx = - \frac{1}{2 x^{2}}$
  So, the result is: $\frac{1}{4 x^{2}}$
Now simplify:

$\frac{- 2 \log{\left(x \right)} - 1}{4 x^{2}}$
Add the constant of integration:

$\frac{- 2 \log{\left(x \right)} - 1}{4 x^{2}}+ \mathrm{constant}$

The answer is:

$\frac{- 2 \log{\left(x \right)} - 1}{4 x^{2}}+ \mathrm{constant}$

The answer (Indefinite) [src]

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

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

Simplify

The answer [src]

-oo

$$-\infty$$

-oo

$$-\infty$$

Упростить

Numerical answer [src]

-3.98309783194748e+39

-3.98309783194748e+39

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

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Identical expressions

Integral of lgx/x^3 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #2