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

The solution

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  8                 
  /                 
 |                  
 |        x         
 |  ------------- dx
 |        _______   
 |  1 + \/ x + 1    
 |                  
/                   
3

$$\int\limits_{3}^{8} \frac{x}{\sqrt{x + 1} + 1}\, dx$$

Integral(x/(1 + sqrt(x + 1)), (x, 3, 8))

Detail solution

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

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

$\int \frac{2 u \left(u^{2} - 1\right)}{u + 1}\, du$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{u \left(u^{2} - 1\right)}{u + 1}\, du = 2 \int \frac{u \left(u^{2} - 1\right)}{u + 1}\, du$
  1. There are multiple ways to do this integral.
    Method #1
    1. Rewrite the integrand:
      
      $\frac{u \left(u^{2} - 1\right)}{u + 1} = u^{2} - u$
    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:
      
      $\int \left(- u\right)\, du = - \int u\, du$
      
      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: $- \frac{u^{2}}{2}$
      
      The result is: $\frac{u^{3}}{3} - \frac{u^{2}}{2}$
    Method #2
    1. Rewrite the integrand:
      
      $\frac{u \left(u^{2} - 1\right)}{u + 1} = \frac{u^{3}}{u + 1} - \frac{u}{u + 1}$
    2. Integrate term-by-term:
      
      Rewrite the integrand:
      
      $\frac{u^{3}}{u + 1} = u^{2} - u + 1 - \frac{1}{u + 1}$
      
      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:
      
      $\int \left(- u\right)\, du = - \int u\, du$
      
      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: $- \frac{u^{2}}{2}$
      
      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:
      
      $\int \left(- \frac{1}{u + 1}\right)\, du = - \int \frac{1}{u + 1}\, du$
      
      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: $- \log{\left(u + 1 \right)}$
      
      The result is: $\frac{u^{3}}{3} - \frac{u^{2}}{2} + u - \log{\left(u + 1 \right)}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{u}{u + 1}\right)\, du = - \int \frac{u}{u + 1}\, du$
      
      Rewrite the integrand:
      
      $\frac{u}{u + 1} = 1 - \frac{1}{u + 1}$
      
      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:
      
      $\int \left(- \frac{1}{u + 1}\right)\, du = - \int \frac{1}{u + 1}\, du$
      
      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: $- \log{\left(u + 1 \right)}$
      
      The result is: $u - \log{\left(u + 1 \right)}$
      
      So, the result is: $- u + \log{\left(u + 1 \right)}$
      
      The result is: $\frac{u^{3}}{3} - \frac{u^{2}}{2}$
  So, the result is: $\frac{2 u^{3}}{3} - u^{2}$
Now substitute $u$ back in:

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

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

23/3

$$\frac{23}{3}$$

23/3

$$\frac{23}{3}$$

23/3

Numerical answer [src]

7.66666666666667

7.66666666666667

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

Other calculators

How to use it?

Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2