What is the integral of x\sqrt[3]{x+3} ?

The integral of x\sqrt[3]{x+3} is 3/7 (x+3)^{7/3}-9/4 (x+3)^{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]{x+3}

\frac{3}{7} (x + 3)^{\frac{7}{3}} - \frac{9}{4} (x + 3)^{\frac{4}{3}} + C

Solution steps

Apply u-substitution

Expand

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

= \frac{3}{7} u^{\frac{7}{3}} - \frac{9}{4} u^{\frac{4}{3}}

Substitute back

u = x + 3

= \frac{3}{7} (x + 3)^{\frac{7}{3}} - \frac{9}{4} (x + 3)^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{7} (x + 3)^{\frac{7}{3}} - \frac{9}{4} (x + 3)^{\frac{4}{3}} + C

What is the integral of x\sqrt[3]{x+3} ?
The integral of x\sqrt[3]{x+3} is 3/7 (x+3)^{7/3}-9/4 (x+3)^{4/3}+C