What is the integral of y/(sqrt(9-y)) ?

The integral of y/(sqrt(9-y)) is -2(-1/3 (9-y)^{3/2}+9sqrt(9-y))+C

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integral of y/(sqrt(9-y))

\int \frac{y}{9 - y} d y

- 2 (- \frac{1}{3} (9 - y)^{\frac{3}{2}} + 9 9 - y) + C

Solution steps

\int \frac{y}{9 - y} d y

Apply u-substitution

: \int - 2 (- u^{2} + 9) d u

= \int - 2 (- u^{2} + 9) 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} + 9 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 9 d u)

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

\int 9 d u = 9 u

= - 2 (- \frac{u ^{3}}{3} + 9 u)

Substitute back

u = 9 - y

= - 2 (- \frac{( 9 - y ) ^{3}}{3} + 9 9 - y)

Simplify

- 2 (- \frac{( 9 - y ) ^{3}}{3} + 9 9 - y) : - 2 (- \frac{1}{3} (9 - y)^{\frac{3}{2}} + 9 9 - y)

= - 2 (- \frac{1}{3} (9 - y)^{\frac{3}{2}} + 9 9 - y)

Add a constant to the solution

= - 2 (- \frac{1}{3} (9 - y)^{\frac{3}{2}} + 9 9 - y) + C

\int 7 x^{2} - 9 d x

\int (x^{2} + cos (x)) d x

\int cos (y) sin (y) + y^{2} d y

\int 4 cos (5 x) d x

\frac{1}{x ^{2} + 1}

What is the integral of y/(sqrt(9-y)) ?
The integral of y/(sqrt(9-y)) is -2(-1/3 (9-y)^{3/2}+9sqrt(9-y))+C