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

The integral of (x^3+1)^{1/3}x^2 is 1/4 (x^3+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 (x^3+1)^{1/3}x^2

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

\frac{1}{4} (x^{3} + 1)^{\frac{4}{3}} + C

Solution steps

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

Apply u-substitution

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

Apply the Power Rule

= \frac{1}{3} \cdot \frac{3}{4} u^{\frac{4}{3}}

Substitute back

u = x^{3} + 1

= \frac{1}{3} \cdot \frac{3}{4} (x^{3} + 1)^{\frac{4}{3}}

Simplify

\frac{1}{3} \cdot \frac{3}{4} (x^{3} + 1)^{\frac{4}{3}} : \frac{1}{4} (x^{3} + 1)^{\frac{4}{3}}

= \frac{1}{4} (x^{3} + 1)^{\frac{4}{3}}

Add a constant to the solution

= \frac{1}{4} (x^{3} + 1)^{\frac{4}{3}} + C

What is the integral of (x^3+1)^{1/3}x^2 ?
The integral of (x^3+1)^{1/3}x^2 is 1/4 (x^3+1)^{4/3}+C