What is the integral of (1/((x-y)^{1/2)}) ?

The integral of (1/((x-y)^{1/2)}) is -2(x-y)^{1/2}+C

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integral of (1/((x-y)^{1/2)})

\int (\frac{1}{( x - y ) ^{\frac{1}{2}}}) d y

- 2 (x - y)^{\frac{1}{2}} + C

Solution steps

\int \frac{1}{( x - y ) ^{\frac{1}{2}}} d y

Apply u-substitution

: \int - \frac{1}{u ^{\frac{1}{2}}} d u

= \int - \frac{1}{u ^{\frac{1}{2}}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{1}{u ^{\frac{1}{2}}} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= - 2 u^{\frac{1}{2}}

Substitute back

u = x - y

= - 2 (x - y)^{\frac{1}{2}}

Add a constant to the solution

= - 2 (x - y)^{\frac{1}{2}} + C

\frac{3}{x + 4}

ln (x - 3)

f (t) = 2 e^{- t} cos (2 π t) - 1

\int 2 1 - x^{2} d x

\frac{d y}{d x}, e^{(\frac{x}{y})} = 6 x - y

What is the integral of (1/((x-y)^{1/2)}) ?
The integral of (1/((x-y)^{1/2)}) is -2(x-y)^{1/2}+C