What is the integral of (x^3)/(\sqrt[3]{x^4+1)} ?

The integral of (x^3)/(\sqrt[3]{x^4+1)} is 3/8 (x^4+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^4+1)}

\frac{3}{8} (x^{4} + 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

Apply the Power Rule

= \frac{1}{4} \cdot \frac{3}{2} u^{\frac{2}{3}}

Substitute back

u = x^{4} + 1

= \frac{1}{4} \cdot \frac{3}{2} (x^{4} + 1)^{\frac{2}{3}}

Simplify

\frac{1}{4} \cdot \frac{3}{2} (x^{4} + 1)^{\frac{2}{3}} : \frac{3}{8} (x^{4} + 1)^{\frac{2}{3}}

= \frac{3}{8} (x^{4} + 1)^{\frac{2}{3}}

Add a constant to the solution

= \frac{3}{8} (x^{4} + 1)^{\frac{2}{3}} + C

What is the integral of (x^3)/(\sqrt[3]{x^4+1)} ?
The integral of (x^3)/(\sqrt[3]{x^4+1)} is 3/8 (x^4+1)^{2/3}+C