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

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

The solution

You have entered [src]

  6                   
  /                   
 |                    
 |         1          
 |  --------------- dx
 |        _________   
 |  1 + \/ 3*x - 2    
 |                    
/                     
1

$$\int\limits_{1}^{6} \frac{1}{\sqrt{3 x - 2} + 1}\, dx$$

Integral(1/(1 + sqrt(3*x - 2)), (x, 1, 6))

Detail solution

Let $u = \sqrt{3 x - 2}$ .

Then let $du = \frac{3 dx}{2 \sqrt{3 x - 2}}$ and substitute $2 du$ :

$\int \frac{2 u}{3 u + 3}\, du$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{u}{3 u + 3}\, du = 2 \int \frac{u}{3 u + 3}\, du$
  1. Rewrite the integrand:
    
    $\frac{u}{3 u + 3} = \frac{1}{3} - \frac{1}{3 \left(u + 1\right)}$
  2. Integrate term-by-term:
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{3}\, du = \frac{u}{3}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{3 \left(u + 1\right)}\right)\, du = - \frac{\int \frac{1}{u + 1}\, du}{3}$
      1. Let $u = u + 1$ .
        
        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:
        
        $\log{\left(u + 1 \right)}$
      So, the result is: $- \frac{\log{\left(u + 1 \right)}}{3}$
    The result is: $\frac{u}{3} - \frac{\log{\left(u + 1 \right)}}{3}$
  So, the result is: $\frac{2 u}{3} - \frac{2 \log{\left(u + 1 \right)}}{3}$
Now substitute $u$ back in:

$\frac{2 \sqrt{3 x - 2}}{3} - \frac{2 \log{\left(\sqrt{3 x - 2} + 1 \right)}}{3}$
Now simplify:

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

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

The answer is:

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

The answer (Indefinite) [src]

  /                                                               
 |                               /      _________\       _________
 |        1                 2*log\1 + \/ 3*x - 2 /   2*\/ 3*x - 2 
 | --------------- dx = C - ---------------------- + -------------
 |       _________                    3                    3      
 | 1 + \/ 3*x - 2                                                 
 |                                                                
/

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

    2*log(5)   2*log(2)
2 - -------- + --------
       3          3

$$- \frac{2 \log{\left(5 \right)}}{3} + \frac{2 \log{\left(2 \right)}}{3} + 2$$

    2*log(5)   2*log(2)
2 - -------- + --------
       3          3

$$- \frac{2 \log{\left(5 \right)}}{3} + \frac{2 \log{\left(2 \right)}}{3} + 2$$

2 - 2*log(5)/3 + 2*log(2)/3

Numerical answer [src]

1.3891395120839

1.3891395120839

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

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

Limits of integration:

The graph:

Piecewise:

The solution