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(one -3x)/(sqrt(x)- two)
(1 minus 3x) divide by ( square root of (x) minus 2)
(one minus 3x) divide by ( square root of (x) minus two)
(1-3x)/(√(x)-2)
1-3x/sqrtx-2
(1-3x) divide by (sqrt(x)-2)
(1-3x)/(sqrt(x)-2)dx
Similar expressions
(1-3x)/(sqrt(x)+2)
(1+3x)/(sqrt(x)-2)

Integral of (1-3x)/(sqrt(x)-2) dx

The solution

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  1             
  /             
 |              
 |   1 - 3*x    
 |  --------- dx
 |    ___       
 |  \/ x  - 2   
 |              
/               
0

$$\int\limits_{0}^{1} \frac{1 - 3 x}{\sqrt{x} - 2}\, dx$$

Integral((1 - 3*x)/(sqrt(x) - 2), (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 $- du$ :
  
  $\int \left(- \frac{6 u^{3} - 2 u}{u - 2}\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{6 u^{3} - 2 u}{u - 2}\, du = - \int \frac{6 u^{3} - 2 u}{u - 2}\, du$
    1. Rewrite the integrand:
      
      $\frac{6 u^{3} - 2 u}{u - 2} = 6 u^{2} + 12 u + 22 + \frac{44}{u - 2}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 6 u^{2}\, du = 6 \int u^{2}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
      
      So, the result is: $2 u^{3}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 12 u\, du = 12 \int u\, du$
      
      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: $6 u^{2}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 22\, du = 22 u$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{44}{u - 2}\, du = 44 \int \frac{1}{u - 2}\, du$
      
      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: $44 \log{\left(u - 2 \right)}$
      
      The result is: $2 u^{3} + 6 u^{2} + 22 u + 44 \log{\left(u - 2 \right)}$
    So, the result is: $- 2 u^{3} - 6 u^{2} - 22 u - 44 \log{\left(u - 2 \right)}$
  Now substitute $u$ back in:
  
  $- 2 x^{\frac{3}{2}} - 22 \sqrt{x} - 6 x - 44 \log{\left(\sqrt{x} - 2 \right)}$
Method #2
1. Rewrite the integrand:
  
  $\frac{1 - 3 x}{\sqrt{x} - 2} = - \frac{3 x - 1}{\sqrt{x} - 2}$
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \frac{3 x - 1}{\sqrt{x} - 2}\right)\, dx = - \int \frac{3 x - 1}{\sqrt{x} - 2}\, dx$
  1. Let $u = \sqrt{x}$ .
    
    Then let $du = \frac{dx}{2 \sqrt{x}}$ and substitute $du$ :
    
    $\int \frac{6 u^{3} - 2 u}{u - 2}\, du$
    1. Rewrite the integrand:
      
      $\frac{6 u^{3} - 2 u}{u - 2} = 6 u^{2} + 12 u + 22 + \frac{44}{u - 2}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 6 u^{2}\, du = 6 \int u^{2}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
      
      So, the result is: $2 u^{3}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 12 u\, du = 12 \int u\, du$
      
      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: $6 u^{2}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 22\, du = 22 u$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{44}{u - 2}\, du = 44 \int \frac{1}{u - 2}\, du$
      
      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: $44 \log{\left(u - 2 \right)}$
      
      The result is: $2 u^{3} + 6 u^{2} + 22 u + 44 \log{\left(u - 2 \right)}$
    Now substitute $u$ back in:
    
    $2 x^{\frac{3}{2}} + 22 \sqrt{x} + 6 x + 44 \log{\left(\sqrt{x} - 2 \right)}$
  So, the result is: $- 2 x^{\frac{3}{2}} - 22 \sqrt{x} - 6 x - 44 \log{\left(\sqrt{x} - 2 \right)}$
Method #3
1. Rewrite the integrand:
  
  $\frac{1 - 3 x}{\sqrt{x} - 2} = - \frac{3 x}{\sqrt{x} - 2} + \frac{1}{\sqrt{x} - 2}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{3 x}{\sqrt{x} - 2}\right)\, dx = - 3 \int \frac{x}{\sqrt{x} - 2}\, dx$
    1. Let $u = \sqrt{x}$ .
      
      Then let $du = \frac{dx}{2 \sqrt{x}}$ and substitute $2 du$ :
      
      $\int \frac{2 u^{3}}{u - 2}\, du$
      
      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: $- 2 x^{\frac{3}{2}} - 24 \sqrt{x} - 6 x - 48 \log{\left(\sqrt{x} - 2 \right)}$
  1. Let $u = \sqrt{x}$ .
    
    Then let $du = \frac{dx}{2 \sqrt{x}}$ and substitute $2 du$ :
    
    $\int \frac{2 u}{u - 2}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u}{u - 2}\, du = 2 \int \frac{u}{u - 2}\, du$
      
      Rewrite the integrand:
      
      $\frac{u}{u - 2} = 1 + \frac{2}{u - 2}$
      
      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:
      
      $\int \frac{2}{u - 2}\, du = 2 \int \frac{1}{u - 2}\, du$
      
      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: $2 \log{\left(u - 2 \right)}$
      
      The result is: $u + 2 \log{\left(u - 2 \right)}$
      
      So, the result is: $2 u + 4 \log{\left(u - 2 \right)}$
    Now substitute $u$ back in:
    
    $2 \sqrt{x} + 4 \log{\left(\sqrt{x} - 2 \right)}$
  The result is: $- 2 x^{\frac{3}{2}} - 22 \sqrt{x} - 6 x - 44 \log{\left(\sqrt{x} - 2 \right)}$
Add the constant of integration:

$- 2 x^{\frac{3}{2}} - 22 \sqrt{x} - 6 x - 44 \log{\left(\sqrt{x} - 2 \right)}+ \mathrm{constant}$

The answer is:

$- 2 x^{\frac{3}{2}} - 22 \sqrt{x} - 6 x - 44 \log{\left(\sqrt{x} - 2 \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                               
 |                                                                
 |  1 - 3*x                 /       ___\        ___            3/2
 | --------- dx = C - 44*log\-2 + \/ x / - 22*\/ x  - 6*x - 2*x   
 |   ___                                                          
 | \/ x  - 2                                                      
 |                                                                
/

$$\int \frac{1 - 3 x}{\sqrt{x} - 2}\, dx = C - 2 x^{\frac{3}{2}} - 22 \sqrt{x} - 6 x - 44 \log{\left(\sqrt{x} - 2 \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

-30 + 44*log(2)

$$-30 + 44 \log{\left(2 \right)}$$

-30 + 44*log(2)

$$-30 + 44 \log{\left(2 \right)}$$

-30 + 44*log(2)

Numerical answer [src]

0.498475944637594

0.498475944637594

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

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Identical expressions

Similar expressions

Integral of (1-3x)/(sqrt(x)-2) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3