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Identical expressions
x* two *(x+y)/(two *x+ one)*dx
x multiply by 2 multiply by (x plus y) divide by (2 multiply by x plus 1) multiply by dx
x multiply by two multiply by (x plus y) divide by (two multiply by x plus one) multiply by dx
x2(x+y)/(2x+1)dx
x2x+y/2x+1dx
x*2*(x+y) divide by (2*x+1)*dx
Similar expressions
x*2*(x-y)/(2*x+1)*dx
x*2*(x+y)/(2*x-1)*dx

Integral of x2(x+y)/(2x+1)dx dx

The solution

You have entered [src]

  1               
  /               
 |                
 |  x*2*(x + y)   
 |  ----------- dx
 |    2*x + 1     
 |                
/                 
0

$$\int\limits_{0}^{1} \frac{2 x \left(x + y\right)}{2 x + 1}\, dx$$

Integral(((x*2)*(x + y))/(2*x + 1), (x, 0, 1))

Detail solution

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

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

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

The answer is:

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

The answer (Indefinite) [src]

  /                                                          
 |                       2                                   
 | x*2*(x + y)          x    x         (1/2 - y)*log(1 + 2*x)
 | ----------- dx = C + -- - - + x*y + ----------------------
 |   2*x + 1            2    2                   2           
 |                                                           
/

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

Simplify

The answer [src]

    (-1 + 2*y)*log(3)
y - -----------------
            4

$$y - \frac{\left(2 y - 1\right) \log{\left(3 \right)}}{4}$$

    (-1 + 2*y)*log(3)
y - -----------------
            4

$$y - \frac{\left(2 y - 1\right) \log{\left(3 \right)}}{4}$$

y - (-1 + 2*y)*log(3)/4

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of x2(x+y)/(2x+1)dx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of x*2*(x+y)/(2*x+1)*dx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Integral of x2(x+y)/(2x+1)dx dx