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

The integral of (x^5)/(\sqrt[3]{1+x^3)} is 1/10 (2x^3-3)(x^3+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^5)/(\sqrt[3]{1+x^3)}

\int \frac{x ^{5}}{3 1 + x ^{3}} d x

\frac{1}{10} (2 x^{3} - 3) (x^{3} + 1)^{\frac{2}{3}} + C

Solution steps

\int \frac{x ^{5}}{3 1 + x ^{3}} d x

Apply u-substitution

= \int \frac{u - 1}{3 3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{u - 1}{3 u} d u

Expand

\frac{u - 1}{3 u} : u^{\frac{2}{3}} - \frac{1}{3 u}

= \frac{1}{3} \cdot \int u^{\frac{2}{3}} - \frac{1}{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}{3} (\int u^{\frac{2}{3}} d u - \int \frac{1}{3 u} d u)

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

\int \frac{1}{3 u} d u = \frac{3}{2} u^{\frac{2}{3}}

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

Substitute back

u = 1 + x^{3}

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

Simplify

\frac{1}{3} (\frac{3}{5} (1 + x^{3})^{\frac{5}{3}} - \frac{3}{2} (1 + x^{3})^{\frac{2}{3}}) : \frac{1}{10} (2 x^{3} - 3) (x^{3} + 1)^{\frac{2}{3}}

= \frac{1}{10} (2 x^{3} - 3) (x^{3} + 1)^{\frac{2}{3}}

Add a constant to the solution

= \frac{1}{10} (2 x^{3} - 3) (x^{3} + 1)^{\frac{2}{3}} + C

\int (tan (2 x) cot (2 x))^{2} d x

\int x^{3} 3^{x^{4}} d x

\int cos (4 x) x d x

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

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

What is the integral of (x^5)/(\sqrt[3]{1+x^3)} ?
The integral of (x^5)/(\sqrt[3]{1+x^3)} is 1/10 (2x^3-3)(x^3+1)^{2/3}+C