What is the integral of (-4x^3-8x^2-4) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (-4x^3-8x^2-4)

\int (- 4 x^{3} - 8 x^{2} - 4) d x

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

Solution steps

\int - 4 x^{3} - 8 x^{2} - 4 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 x^{3} d x - \int 8 x^{2} d x - \int 4 d x

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

\int 8 x^{2} d x = \frac{8 x ^{3}}{3}

\int 4 d x = 4 x

= - x^{4} - \frac{8 x ^{3}}{3} - 4 x

Add a constant to the solution

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

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

g (t) = 5 (cos (π t))^{2}

\int - 6 sin (3 x) d x

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

(2, - 4), (6, 0)

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