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The solution

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

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

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

Detail solution

Let $u = x - y$ .

Then let $du = - dy$ and substitute $- du$ :

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

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The answer [src]

-log(1 - x) + log(-x)

$$\log{\left(- x \right)} - \log{\left(1 - x \right)}$$

-log(1 - x) + log(-x)

$$\log{\left(- x \right)} - \log{\left(1 - x \right)}$$

-log(1 - x) + log(-x)

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

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Integral of 1/(x-y) dl

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The solution