What is the integral of+(y+3)/((3-y)^{2/3)} ?

The integral of+(y+3)/((3-y)^{2/3)} is 3/4 (3-y)^{4/3}-18(3-y)^{1/3}+C

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integral of+(y+3)/((3-y)^{2/3)}

\int + \frac{y + 3}{( 3 - y ) ^{\frac{2}{3}}} d y

\frac{3}{4} (3 - y)^{\frac{4}{3}} - 18 (3 - y)^{\frac{1}{3}} + C

Solution steps

\int \frac{y + 3}{( 3 - y ) ^{\frac{2}{3}}} d y

Apply u-substitution

: \int - \frac{- u + 6}{u ^{\frac{2}{3}}} d u

= \int - \frac{- u + 6}{u ^{\frac{2}{3}}} d u

Take the constant out:

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

= - \int \frac{- u + 6}{u ^{\frac{2}{3}}} d u

Expand

\frac{- u + 6}{u ^{\frac{2}{3}}} : - u^{\frac{1}{3}} + \frac{6}{u ^{\frac{2}{3}}}

= - \int - u^{\frac{1}{3}} + \frac{6}{u ^{\frac{2}{3}}} 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

= - (- \int u^{\frac{1}{3}} d u + \int \frac{6}{u ^{\frac{2}{3}}} d u)

\int u^{\frac{1}{3}} d u = \frac{3}{4} u^{\frac{4}{3}}

\int \frac{6}{u ^{\frac{2}{3}}} d u = 18 u^{\frac{1}{3}}

= - (- \frac{3}{4} u^{\frac{4}{3}} + 18 u^{\frac{1}{3}})

Substitute back

u = 3 - y

= - (- \frac{3}{4} (3 - y)^{\frac{4}{3}} + 18 (3 - y)^{\frac{1}{3}})

Simplify

= \frac{3}{4} (3 - y)^{\frac{4}{3}} - 18 (3 - y)^{\frac{1}{3}}

Add a constant to the solution

= \frac{3}{4} (3 - y)^{\frac{4}{3}} - 18 (3 - y)^{\frac{1}{3}} + C

(4 + x) \frac{d y}{d x} = 9 y

\int - 7 e^{- x} d x

\int \frac{1}{x ^{2} + 2 x + 50} d x

x \to 1 + lim (\frac{2 + x}{1 - x})

x \to 3 lim (2 - \frac{1}{x})

What is the integral of+(y+3)/((3-y)^{2/3)} ?
The integral of+(y+3)/((3-y)^{2/3)} is 3/4 (3-y)^{4/3}-18(3-y)^{1/3}+C