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Identical expressions
(lnxdx)/(x(one +lnx)^ one / two)
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(lnxdx) divide by (x(one plus lnx) to the power of one divide by two)
(lnxdx)/(x(1+lnx)1/2)
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Similar expressions
(lnxdx)/(x(1-lnx)^1/2)

Solving integrals/ (lnxdx)/(x(1+lnx)^1/2)

Integral of (lnxdx)/(x(1+lnx)^1/2) dx

The solution

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  1                    
  /                    
 |                     
 |       log(x)        
 |  ---------------- dx
 |      ____________   
 |  x*\/ 1 + log(x)    
 |                     
/                      
0

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

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

$\frac{2 \left(\log{\left(x \right)} - 2\right) \sqrt{\log{\left(x \right)} + 1}}{3}$
Add the constant of integration:

$\frac{2 \left(\log{\left(x \right)} - 2\right) \sqrt{\log{\left(x \right)} + 1}}{3}+ \mathrm{constant}$

The answer is:

$\frac{2 \left(\log{\left(x \right)} - 2\right) \sqrt{\log{\left(x \right)} + 1}}{3}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                              
 |                                                            3/2
 |      log(x)                   ____________   2*(1 + log(x))   
 | ---------------- dx = C - 2*\/ 1 + log(x)  + -----------------
 |     ____________                                     3        
 | x*\/ 1 + log(x)                                               
 |                                                               
/

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

-4/3 + oo*I

$$- \frac{4}{3} + \infty i$$

-4/3 + oo*I

$$- \frac{4}{3} + \infty i$$

-4/3 + oo*i

Numerical answer [src]

(-1.0818887836703 + 202.533724926781j)

(-1.0818887836703 + 202.533724926781j)

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

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Integral of (lnxdx)/(x(1+lnx)^1/2) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #2