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Identical expressions
(x+xy)dy/(yx-y)
(x plus xy)dy divide by (yx minus y)
x+xydy/yx-y
(x+xy)dy divide by (yx-y)
(x+xy)dy/(yx-y)dx
Similar expressions
(x+xy)dy/(yx+y)
(x-xy)dy/(yx-y)

Integral of (x+xy)dy/(yx-y) dx

The solution

You have entered [src]

  1           
  /           
 |            
 |  x + x*y   
 |  ------- dy
 |  y*x - y   
 |            
/             
0

$$\int\limits_{0}^{1} \frac{x y + x}{x y - y}\, dy$$

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

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\frac{x y + x}{x y - y} = \frac{x}{x - 1} + \frac{x}{y \left(x - 1\right)}$
2. Integrate term-by-term:
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \frac{x}{x - 1}\, dy = \frac{x y}{x - 1}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{x}{y \left(x - 1\right)}\, dy = \frac{x \int \frac{1}{y}\, dy}{x - 1}$
    1. The integral of $\frac{1}{y}$ is $\log{\left(y \right)}$ .
    So, the result is: $\frac{x \log{\left(y \right)}}{x - 1}$
  The result is: $\frac{x y}{x - 1} + \frac{x \log{\left(y \right)}}{x - 1}$
Method #2
1. Rewrite the integrand:
  
  $\frac{x y + x}{x y - y} = \frac{x y}{x y - y} + \frac{x}{x y - y}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{x y}{x y - y}\, dy = x \int \frac{y}{x y - y}\, dy$
    1. Rewrite the integrand:
      
      $\frac{y}{x y - y} = \frac{1}{x - 1}$
    2. The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{x - 1}\, dy = \frac{y}{x - 1}$
    So, the result is: $\frac{x y}{x - 1}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{x}{x y - y}\, dy = x \int \frac{1}{x y - y}\, dy$
    1. Rewrite the integrand:
      
      $\frac{1}{x y - y} = \frac{1}{y \left(x - 1\right)}$
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{y \left(x - 1\right)}\, dy = \frac{\int \frac{1}{y}\, dy}{x - 1}$
      
      The integral of $\frac{1}{y}$ is $\log{\left(y \right)}$ .
      
      So, the result is: $\frac{\log{\left(y \right)}}{x - 1}$
    So, the result is: $\frac{x \log{\left(y \right)}}{x - 1}$
  The result is: $\frac{x y}{x - 1} + \frac{x \log{\left(y \right)}}{x - 1}$
Now simplify:

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The answer [src]

       /  x   \     x   
oo*sign|------| + ------
       \-1 + x/   -1 + x

$$\frac{x}{x - 1} + \infty \operatorname{sign}{\left(\frac{x}{x - 1} \right)}$$

       /  x   \     x   
oo*sign|------| + ------
       \-1 + x/   -1 + x

$$\frac{x}{x - 1} + \infty \operatorname{sign}{\left(\frac{x}{x - 1} \right)}$$

oo*sign(x/(-1 + x)) + x/(-1 + x)

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

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Identical expressions

Similar expressions

Integral of (x+xy)dy/(yx-y) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2