What is the integral of (4e^x-x^3+6x-8) ?

The integral of (4e^x-x^3+6x-8) is 4e^x-(x^4)/4+3x^2-8x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (4e^x-x^3+6x-8)

\int (4 e^{x} - x^{3} + 6 x - 8) d x

4 e^{x} - \frac{x ^{4}}{4} + 3 x^{2} - 8 x + C

Solution steps

\int 4 e^{x} - x^{3} + 6 x - 8 d x

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 e^{x} d x - \int x^{3} d x + \int 6 x d x - \int 8 d x

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

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

\int 6 x d x = 3 x^{2}

\int 8 d x = 8 x

= 4 e^{x} - \frac{x ^{4}}{4} + 3 x^{2} - 8 x

Add a constant to the solution

= 4 e^{x} - \frac{x ^{4}}{4} + 3 x^{2} - 8 x + C

n = 0 \sum \infty \frac{1}{4 n + 1}

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

y^{^{''}} + 4 y^{^{'}} + 2 y = 0

\frac{\partial}{\partial u} (tan (\frac{u}{v}))

\int \frac{x}{( 3 x ^{2} - 4 ) ^{2}} d x

What is the integral of (4e^x-x^3+6x-8) ?
The integral of (4e^x-x^3+6x-8) is 4e^x-(x^4)/4+3x^2-8x+C