What is the integral of x\sqrt[3]{5-2x} ?

The integral of x\sqrt[3]{5-2x} is 1/4 (3/7 (5-2x)^{7/3}-15/4 (5-2x)^{4/3})+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of x\sqrt[3]{5-2x}

\int x 3 5 - 2 x d x

\frac{1}{4} (\frac{3}{7} (5 - 2 x)^{\frac{7}{3}} - \frac{15}{4} (5 - 2 x)^{\frac{4}{3}}) + C

Solution steps

\int x 3 5 - 2 x d x

Apply u-substitution

= \int \frac{3 u ( u - 5 )}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int 3 u (u - 5) d u

Expand

3 u (u - 5) : u^{\frac{4}{3}} - 5 3 u

= \frac{1}{4} \cdot \int u^{\frac{4}{3}} - 5 3 u 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

= \frac{1}{4} (\int u^{\frac{4}{3}} d u - \int 5 3 u d u)

\int u^{\frac{4}{3}} d u = \frac{3}{7} u^{\frac{7}{3}}

\int 5 3 u d u = \frac{15}{4} u^{\frac{4}{3}}

= \frac{1}{4} (\frac{3}{7} u^{\frac{7}{3}} - \frac{15}{4} u^{\frac{4}{3}})

Substitute back

u = 5 - 2 x

= \frac{1}{4} (\frac{3}{7} (5 - 2 x)^{\frac{7}{3}} - \frac{15}{4} (5 - 2 x)^{\frac{4}{3}})

Add a constant to the solution

= \frac{1}{4} (\frac{3}{7} (5 - 2 x)^{\frac{7}{3}} - \frac{15}{4} (5 - 2 x)^{\frac{4}{3}}) + C

\int (x^{3} + 1) ln (x) d x

\int x^{3} a^{2} + x^{2} d x

\int y^{2} sin (2 y) d y

\int \frac{e ^{2 x}}{9 - 4 e ^{4 x}} d x

\int x (\frac{- 3}{x ^{3}} - x^{2} + \frac{x}{4}) d x

What is the integral of x\sqrt[3]{5-2x} ?
The integral of x\sqrt[3]{5-2x} is 1/4 (3/7 (5-2x)^{7/3}-15/4 (5-2x)^{4/3})+C