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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

= 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

\int \frac{4 x ^{10} - 2 x ^{4} + 15 x ^{2}}{x ^{3}} d x

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

\int x^{\frac{- 7}{8}} d x

\int 6 x^{2} (2 x^{3} + 5)^{5} d x

\int \frac{2 x}{y} - \frac{3 y ^{2}}{x ^{4}} d x

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