What is the integral of (2r)/((1-r)^{2/3)} ?

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

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integral of (2r)/((1-r)^{2/3)}

\int \frac{2 r}{( 1 - r ) ^{\frac{2}{3}}} d r

- 2 (- \frac{3}{4} (1 - r)^{\frac{4}{3}} + 3 (1 - r)^{\frac{1}{3}}) + C

Solution steps

\int \frac{2 r}{( 1 - r ) ^{\frac{2}{3}}} d r

Take the constant out:

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

= 2 \cdot \int \frac{r}{( 1 - r ) ^{\frac{2}{3}}} d r

Apply u-substitution

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

Take the constant out:

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

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

Expand

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

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

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

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

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

= 2 (- (- \frac{3}{4} u^{\frac{4}{3}} + 3 u^{\frac{1}{3}}))

Substitute back

u = 1 - r

= 2 (- (- \frac{3}{4} (1 - r)^{\frac{4}{3}} + 3 (1 - r)^{\frac{1}{3}}))

Simplify

= - 2 (- \frac{3}{4} (1 - r)^{\frac{4}{3}} + 3 (1 - r)^{\frac{1}{3}})

Add a constant to the solution

= - 2 (- \frac{3}{4} (1 - r)^{\frac{4}{3}} + 3 (1 - r)^{\frac{1}{3}}) + C

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