What is the integral of (6t^2*\sqrt[3]{t}) ?

The integral of (6t^2*\sqrt[3]{t}) is 6(1/3 t^3\sqrt[3]{t}-1/30 t^{10/3})+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (6t^2*\sqrt[3]{t})

Solution steps

Take the constant out:

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

Apply Integration By Parts

\int \frac{1}{9} t^{\frac{7}{3}} d t = \frac{1}{30} t^{\frac{10}{3}}

Add a constant to the solution

\frac{d}{d x} (tan (x^{2}))

\frac{\partial}{\partial y} (x - y^{2})

\frac{d}{d x} ((- cos (4 x) + 1)^{2} + 1)

y = - x^{2} + 4, y = 3 x

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

What is the integral of (6t^2*\sqrt[3]{t}) ?
The integral of (6t^2*\sqrt[3]{t}) is 6(1/3 t^3\sqrt[3]{t}-1/30 t^{10/3})+C