What is the integral of 1/(sqrt(y-x)) ?

The integral of 1/(sqrt(y-x)) is 2sqrt(-x+y)+C

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integral of 1/(sqrt(y-x))

\int \frac{1}{y - x} d y

2 - x + y + C

Solution steps

\int \frac{1}{y - x} d y

Apply u-substitution

: \int \frac{1}{u} d u

= \int \frac{1}{u} 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}}

Apply radical rule:

= 2 - x + y

Add a constant to the solution

= 2 - x + y + C

\int (1. 8^{0.045 x} - 1) d x

f (x) = e^{\frac{x}{3}}

t^{2} y^{^{''}} + 4 t y^{^{'}} + 2 y = 0

x \to 0 + lim ((x^{2}) ln (x))

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