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

y = \frac{1}{x ^{2} + 1}, x = 0, x = 6, y = 0

\frac{d}{d x} (- 6 x^{2} + 5)

\frac{d}{d x} (arccos (x - 1))

\int_{- 15}^{0} \frac{1}{1 - x} d x

3 e^{- y} (x^{2} + y^{2}) + 4

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