What is the integral of 4x^3-\sqrt[3]{x}+x^e ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 4x^3-\sqrt[3]{x}+x^e

x^{4} - \frac{3}{4} x^{\frac{4}{3}} + \frac{x ^{e + 1}}{e + 1} + C

Solution steps

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int x^{e} d x = \frac{x ^{e + 1}}{e + 1}

= x^{4} - \frac{3}{4} x^{\frac{4}{3}} + \frac{x ^{e + 1}}{e + 1}

Add a constant to the solution

= x^{4} - \frac{3}{4} x^{\frac{4}{3}} + \frac{x ^{e + 1}}{e + 1} + C

\frac{d}{d x} (ln (x e^{- 9 x}))

\frac{\partial}{\partial x} (\frac{c - b}{a - x})

\frac{d}{d x} (m x^{n} - n x^{m})

y^{^{'}} = \frac{( 1 + x )}{x y ^{14}}

y = (x^{2} - 15)^{4}, at x = 4

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