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Identical expressions
one /(sqrt(3x+ one)+ two)
1 divide by ( square root of (3x plus 1) plus 2)
one divide by ( square root of (3x plus one) plus two)
1/(√(3x+1)+2)
1/sqrt3x+1+2
1 divide by (sqrt(3x+1)+2)
1/(sqrt(3x+1)+2)dx
Similar expressions
1/(sqrt(3x-1)+2)
1/(sqrt(3x+1)-2)

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

The solution

You have entered [src]

  1                   
  /                   
 |                    
 |         1          
 |  --------------- dx
 |    _________       
 |  \/ 3*x + 1  + 2   
 |                    
/                     
0

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

Integral(1/(sqrt(3*x + 1) + 2), (x, 0, 1))

Detail solution

Let $u = \sqrt{3 x + 1}$ .

Then let $du = \frac{3 dx}{2 \sqrt{3 x + 1}}$ and substitute $2 du$ :

$\int \frac{2 u}{3 u + 6}\, du$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{u}{3 u + 6}\, du = 2 \int \frac{u}{3 u + 6}\, du$
  1. Rewrite the integrand:
    
    $\frac{u}{3 u + 6} = \frac{1}{3} - \frac{2}{3 \left(u + 2\right)}$
  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:
      
      $\int \left(- \frac{2}{3 \left(u + 2\right)}\right)\, du = - \frac{2 \int \frac{1}{u + 2}\, du}{3}$
      1. 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: $- \frac{2 \log{\left(u + 2 \right)}}{3}$
    The result is: $\frac{u}{3} - \frac{2 \log{\left(u + 2 \right)}}{3}$
  So, the result is: $\frac{2 u}{3} - \frac{4 \log{\left(u + 2 \right)}}{3}$
Now substitute $u$ back in:

$\frac{2 \sqrt{3 x + 1}}{3} - \frac{4 \log{\left(\sqrt{3 x + 1} + 2 \right)}}{3}$
Now simplify:

$\frac{2 \sqrt{3 x + 1}}{3} - \frac{4 \log{\left(\sqrt{3 x + 1} + 2 \right)}}{3}$
Add the constant of integration:

$\frac{2 \sqrt{3 x + 1}}{3} - \frac{4 \log{\left(\sqrt{3 x + 1} + 2 \right)}}{3}+ \mathrm{constant}$

The answer is:

$\frac{2 \sqrt{3 x + 1}}{3} - \frac{4 \log{\left(\sqrt{3 x + 1} + 2 \right)}}{3}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                               
 |                               /      _________\       _________
 |        1                 4*log\2 + \/ 3*x + 1 /   2*\/ 3*x + 1 
 | --------------- dx = C - ---------------------- + -------------
 |   _________                        3                    3      
 | \/ 3*x + 1  + 2                                                
 |                                                                
/

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

2   4*log(4)   4*log(3)
- - -------- + --------
3      3          3

$$- \frac{4 \log{\left(4 \right)}}{3} + \frac{2}{3} + \frac{4 \log{\left(3 \right)}}{3}$$

2   4*log(4)   4*log(3)
- - -------- + --------
3      3          3

$$- \frac{4 \log{\left(4 \right)}}{3} + \frac{2}{3} + \frac{4 \log{\left(3 \right)}}{3}$$

2/3 - 4*log(4)/3 + 4*log(3)/3

Numerical answer [src]

0.283090570064292

0.283090570064292

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

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

Limits of integration:

The graph:

Piecewise:

The solution