What is the integral of (x^2)/(\sqrt[3]{x+2)} ?

The integral of (x^2)/(\sqrt[3]{x+2)} is 3/8 (x+2)^{8/3}-12/5 (x+2)^{5/3}+6(x+2)^{2/3}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2)/(\sqrt[3]{x+2)}

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

\frac{3}{8} (x + 2)^{\frac{8}{3}} - \frac{12}{5} (x + 2)^{\frac{5}{3}} + 6 (x + 2)^{\frac{2}{3}} + C

Solution steps

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

Apply u-substitution

: \int \frac{( u - 2 ) ^{2}}{3 u} d u

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

Expand

\frac{( u - 2 ) ^{2}}{3 u} : u^{\frac{5}{3}} - 4 u^{\frac{2}{3}} + \frac{4}{3 u}

= \int u^{\frac{5}{3}} - 4 u^{\frac{2}{3}} + \frac{4}{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

= \int u^{\frac{5}{3}} d u - \int 4 u^{\frac{2}{3}} d u + \int \frac{4}{3 u} d u

\int u^{\frac{5}{3}} d u = \frac{3}{8} u^{\frac{8}{3}}

\int 4 u^{\frac{2}{3}} d u = \frac{12}{5} u^{\frac{5}{3}}

\int \frac{4}{3 u} d u = 6 u^{\frac{2}{3}}

= \frac{3}{8} u^{\frac{8}{3}} - \frac{12}{5} u^{\frac{5}{3}} + 6 u^{\frac{2}{3}}

Substitute back

u = x + 2

= \frac{3}{8} (x + 2)^{\frac{8}{3}} - \frac{12}{5} (x + 2)^{\frac{5}{3}} + 6 (x + 2)^{\frac{2}{3}}

Add a constant to the solution

= \frac{3}{8} (x + 2)^{\frac{8}{3}} - \frac{12}{5} (x + 2)^{\frac{5}{3}} + 6 (x + 2)^{\frac{2}{3}} + C

\int - \frac{2}{81 - 49 t ^{2}} d t

\frac{d}{d x} (\frac{x ^{6}}{x ^{2}})

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

y = \frac{1}{x ^{2} + 1}

f (x) = 9 t^{2} + t^{3}

What is the integral of (x^2)/(\sqrt[3]{x+2)} ?
The integral of (x^2)/(\sqrt[3]{x+2)} is 3/8 (x+2)^{8/3}-12/5 (x+2)^{5/3}+6(x+2)^{2/3}+C