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Integral of x/(1-sqrt(x)+1) dx

The solution

You have entered [src]

  1                 
  /                 
 |                  
 |        x         
 |  ------------- dx
 |        ___       
 |  1 - \/ x  + 1   
 |                  
/                   
0

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

Integral(x/(1 - sqrt(x) + 1), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = \sqrt{x}$ .
  
  Then let $du = \frac{dx}{2 \sqrt{x}}$ and substitute $- 2 du$ :
  
  $\int \left(- \frac{2 u^{3}}{u - 2}\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{u^{3}}{u - 2}\, du = - 2 \int \frac{u^{3}}{u - 2}\, du$
    1. Rewrite the integrand:
      
      $\frac{u^{3}}{u - 2} = u^{2} + 2 u + 4 + \frac{8}{u - 2}$
    2. Integrate term-by-term:
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 u\, du = 2 \int u\, du$
      
      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: $u^{2}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 4\, du = 4 u$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{8}{u - 2}\, du = 8 \int \frac{1}{u - 2}\, du$
      
      Let $u = u - 2$ .
      
      Then let $du = du$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(u - 2 \right)}$
      
      So, the result is: $8 \log{\left(u - 2 \right)}$
      
      The result is: $\frac{u^{3}}{3} + u^{2} + 4 u + 8 \log{\left(u - 2 \right)}$
    So, the result is: $- \frac{2 u^{3}}{3} - 2 u^{2} - 8 u - 16 \log{\left(u - 2 \right)}$
  Now substitute $u$ back in:
  
  $- \frac{2 x^{\frac{3}{2}}}{3} - 8 \sqrt{x} - 2 x - 16 \log{\left(\sqrt{x} - 2 \right)}$
Method #2
1. Rewrite the integrand:
  
  $\frac{x}{\left(1 - \sqrt{x}\right) + 1} = - \frac{x}{\sqrt{x} - 2}$
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \frac{x}{\sqrt{x} - 2}\right)\, dx = - \int \frac{x}{\sqrt{x} - 2}\, dx$
  1. Let $u = \sqrt{x}$ .
    
    Then let $du = \frac{dx}{2 \sqrt{x}}$ and substitute $2 du$ :
    
    $\int \frac{2 u^{3}}{u - 2}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u^{3}}{u - 2}\, du = 2 \int \frac{u^{3}}{u - 2}\, du$
      
      Rewrite the integrand:
      
      $\frac{u^{3}}{u - 2} = u^{2} + 2 u + 4 + \frac{8}{u - 2}$
      
      Integrate term-by-term:
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 u\, du = 2 \int u\, du$
      
      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: $u^{2}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 4\, du = 4 u$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{8}{u - 2}\, du = 8 \int \frac{1}{u - 2}\, du$
      
      Let $u = u - 2$ .
      
      Then let $du = du$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(u - 2 \right)}$
      
      So, the result is: $8 \log{\left(u - 2 \right)}$
      
      The result is: $\frac{u^{3}}{3} + u^{2} + 4 u + 8 \log{\left(u - 2 \right)}$
      
      So, the result is: $\frac{2 u^{3}}{3} + 2 u^{2} + 8 u + 16 \log{\left(u - 2 \right)}$
    Now substitute $u$ back in:
    
    $\frac{2 x^{\frac{3}{2}}}{3} + 8 \sqrt{x} + 2 x + 16 \log{\left(\sqrt{x} - 2 \right)}$
  So, the result is: $- \frac{2 x^{\frac{3}{2}}}{3} - 8 \sqrt{x} - 2 x - 16 \log{\left(\sqrt{x} - 2 \right)}$
Add the constant of integration:

$- \frac{2 x^{\frac{3}{2}}}{3} - 8 \sqrt{x} - 2 x - 16 \log{\left(\sqrt{x} - 2 \right)}+ \mathrm{constant}$

The answer is:

$- \frac{2 x^{\frac{3}{2}}}{3} - 8 \sqrt{x} - 2 x - 16 \log{\left(\sqrt{x} - 2 \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                  
 |                                                                3/2
 |       x                      /       ___\       ___         2*x   
 | ------------- dx = C - 16*log\-2 + \/ x / - 8*\/ x  - 2*x - ------
 |       ___                                                     3   
 | 1 - \/ x  + 1                                                     
 |                                                                   
/

$$\int \frac{x}{\left(1 - \sqrt{x}\right) + 1}\, dx = C - \frac{2 x^{\frac{3}{2}}}{3} - 8 \sqrt{x} - 2 x - 16 \log{\left(\sqrt{x} - 2 \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

-32/3 + 16*log(2)

$$- \frac{32}{3} + 16 \log{\left(2 \right)}$$

-32/3 + 16*log(2)

$$- \frac{32}{3} + 16 \log{\left(2 \right)}$$

-32/3 + 16*log(2)

Numerical answer [src]

0.423688222292458

0.423688222292458

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

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Identical expressions

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Integral of x/(1-sqrt(x)+1) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2