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

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

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

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

- 2 b 1 - y + C

Solution steps

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

Take the constant out:

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

= b \cdot \int \frac{1}{1 - y} d y

Apply u-substitution

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

= b \cdot \int - \frac{1}{u} d u

Take the constant out:

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

= b (- \int \frac{1}{u} d u)

Apply the Power Rule

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

= b (- 2 u^{\frac{1}{2}})

Substitute back

u = 1 - y

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

Simplify

b (- 2 (1 - y)^{\frac{1}{2}}) : - 2 b 1 - y

= - 2 b 1 - y

Add a constant to the solution

= - 2 b 1 - y + C

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