What is the integral of (x^3)/(\sqrt[3]{x^2+1)} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x^3)/(\sqrt[3]{x^2+1)}

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

Solution steps

Apply u-substitution

Take the constant out:

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

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{2}{3}} d u = \frac{3}{5} u^{\frac{5}{3}}

= \frac{1}{2} (\frac{3}{5} u^{\frac{5}{3}} - \frac{3}{2} u^{\frac{2}{3}})

Substitute back

u = x^{2} + 1

= \frac{1}{2} (\frac{3}{5} (x^{2} + 1)^{\frac{5}{3}} - \frac{3}{2} (x^{2} + 1)^{\frac{2}{3}})

Add a constant to the solution

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

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