What is the integral of y/(sqrt(6-y)) ?

The integral of y/(sqrt(6-y)) is -2(-1/3 (6-y)^{3/2}+6sqrt(6-y))+C

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integral of y/(sqrt(6-y))

\int \frac{y}{6 - y} d y

- 2 (- \frac{1}{3} (6 - y)^{\frac{3}{2}} + 6 6 - y) + C

Solution steps

\int \frac{y}{6 - y} d y

Apply u-substitution

= \int - 2 (- u^{2} + 6) d u

Take the constant out:

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

= - 2 \cdot \int - u^{2} + 6 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - 2 (- \int u^{2} d u + \int 6 d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int 6 d u = 6 u

= - 2 (- \frac{u ^{3}}{3} + 6 u)

Substitute back

u = 6 - y

= - 2 (- \frac{( 6 - y ) ^{3}}{3} + 6 6 - y)

Simplify

- 2 (- \frac{( 6 - y ) ^{3}}{3} + 6 6 - y) : - 2 (- \frac{1}{3} (6 - y)^{\frac{3}{2}} + 6 6 - y)

= - 2 (- \frac{1}{3} (6 - y)^{\frac{3}{2}} + 6 6 - y)

Add a constant to the solution

= - 2 (- \frac{1}{3} (6 - y)^{\frac{3}{2}} + 6 6 - y) + C

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What is the integral of y/(sqrt(6-y)) ?
The integral of y/(sqrt(6-y)) is -2(-1/3 (6-y)^{3/2}+6sqrt(6-y))+C