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

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

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

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

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

Solution steps

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

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 - x

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

Simplify

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

Add a constant to the solution

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

\int (\frac{ln ( x ^{2} )}{x}) d x

\int \frac{1}{( x + 2 ) ^{5}} d x

\int e^{- x^{2}} x^{2} d x

cos (3 x)

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