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Solving integrals/ 1/(x-y)

Integral of 1/(x-y) dl

Limits of integration:

from to
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Piecewise:

The solution

You have entered [src] 
  1         
  /         
 |          
 |    1     
 |  ----- dy
 |  x - y   
 |          
/           
0
∫011x−y dy\int\limits_{0}^{1} \frac{1}{x - y}\, dy 
Integral(1/(x - y), (y, 0, 1))
Detail solution

  1. Let u=x−yu = x - y.

    Then let du=−dydu = - dy and substitute −du- du:

    ∫(−1u) 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:

      ∫1u du=−∫1u du\int \frac{1}{u}\, du = - \int \frac{1}{u}\, du

      1. The integral of 1u\frac{1}{u} is log⁡(u)\log{\left(u \right)}.

      So, the result is: −log⁡(u)- \log{\left(u \right)}

    Now substitute uu back in:

    −log⁡(x−y)- \log{\left(x - y \right)}

  2. Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src] 
  /                         
 |                          
 |   1                      
 | ----- dy = C - log(x - y)
 | x - y                    
 |                          
/
∫1x−y dy=C−log⁡(x−y)\int \frac{1}{x - y}\, dy = C - \log{\left(x - y \right)}
Simplify
The answer [src] 
-log(1 - x) + log(-x)
log⁡(−x)−log⁡(1−x)\log{\left(- x \right)} - \log{\left(1 - x \right)} 
=
= 
-log(1 - x) + log(-x)
log⁡(−x)−log⁡(1−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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