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

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

\int \frac{x ^{2}}{3 x ^{3} - 1} d x

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

Solution steps

\int \frac{x ^{2}}{3 x ^{3} - 1} d x

Apply u-substitution

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

Apply u-substitution

= \frac{1}{3} \cdot \int \frac{1}{3 v} d v

Apply the Power Rule

= \frac{1}{3} \cdot \frac{3}{2} v^{\frac{2}{3}}

Substitute back

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

Simplify

\frac{1}{3} \cdot \frac{3}{2} (x^{3} - 1)^{\frac{2}{3}} : \frac{1}{2} (x^{3} - 1)^{\frac{2}{3}}

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

Add a constant to the solution

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

\int \frac{1}{x ( x ^{2} + 3 ) ^{2}} d x

\int 2 x^{3} sin (x^{2}) d x

\int (3 - 2 sin (x))^{2} d x

\int \frac{6 x ^{4} + 2 x ^{2} + 4 x + 15}{x ^{2} + 3} d x

\int \frac{2 t ^{2} - t + 4}{t ^{3} + 4 t} d t

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