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Solving integrals/ (1/x-y)dy

Integral of (1/x-y)dy dx

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The solution

You have entered [src] 
  1           
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0
∫01(−y+1x) dy\int\limits_{0}^{1} \left(- y + \frac{1}{x}\right)\, dy 
Integral(1/x - y, (y, 0, 1))
Detail solution

  1. Integrate term-by-term:

    1. The integral of a constant times a function is the constant times the integral of the function:

      ∫(−y) dy=−∫y dy\int \left(- y\right)\, dy = - \int y\, dy

      1. The integral of yny^{n} is yn+1n+1\frac{y^{n + 1}}{n + 1} when n≠−1n \neq -1:

        ∫y dy=y22\int y\, dy = \frac{y^{2}}{2}

      So, the result is: −y22- \frac{y^{2}}{2}

    1. The integral of a constant is the constant times the variable of integration:

      ∫1x dy=yx\int \frac{1}{x}\, dy = \frac{y}{x}

    The result is: −y22+yx- \frac{y^{2}}{2} + \frac{y}{x}

  2. Add the constant of integration:

    −y22+yx+constant- \frac{y^{2}}{2} + \frac{y}{x}+ \mathrm{constant}

The answer is:

−y22+yx+constant- \frac{y^{2}}{2} + \frac{y}{x}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                       
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 | /1    \          y    y
 | |- - y| dy = C - -- + -
 | \x    /          2    x
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/
∫(−y+1x) dy=C−y22+yx\int \left(- y + \frac{1}{x}\right)\, dy = C - \frac{y^{2}}{2} + \frac{y}{x}
Simplify
The answer [src] 
  1   1
- - + -
  2   x
−12+1x- \frac{1}{2} + \frac{1}{x} 
=
= 
  1   1
- - + -
  2   x
−12+1x- \frac{1}{2} + \frac{1}{x} 
-1/2 + 1/x

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

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