What is the integral of y^2\sqrt[3]{y+1} ?

The integral of y^2\sqrt[3]{y+1} is 3/10 (y+1)^{10/3}-6/7 (y+1)^{7/3}+3/4 (y+1)^{4/3}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of y^2\sqrt[3]{y+1}

\frac{3}{10} (y + 1)^{\frac{10}{3}} - \frac{6}{7} (y + 1)^{\frac{7}{3}} + \frac{3}{4} (y + 1)^{\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{7}{3}} d u = \frac{3}{10} u^{\frac{10}{3}}

\int 2 u^{\frac{4}{3}} d u = \frac{6}{7} u^{\frac{7}{3}}

= \frac{3}{10} u^{\frac{10}{3}} - \frac{6}{7} u^{\frac{7}{3}} + \frac{3}{4} u^{\frac{4}{3}}

Substitute back

u = y + 1

= \frac{3}{10} (y + 1)^{\frac{10}{3}} - \frac{6}{7} (y + 1)^{\frac{7}{3}} + \frac{3}{4} (y + 1)^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{10} (y + 1)^{\frac{10}{3}} - \frac{6}{7} (y + 1)^{\frac{7}{3}} + \frac{3}{4} (y + 1)^{\frac{4}{3}} + C

What is the integral of y^2\sqrt[3]{y+1} ?
The integral of y^2\sqrt[3]{y+1} is 3/10 (y+1)^{10/3}-6/7 (y+1)^{7/3}+3/4 (y+1)^{4/3}+C