What is the integral of x(x^2-1)^3 ?

The integral of x(x^2-1)^3 is (x^8)/8-(x^6)/2+(3x^4)/4-(x^2)/2+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of x(x^2-1)^3

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

\frac{x ^{8}}{8} - \frac{x ^{6}}{2} + \frac{3 x ^{4}}{4} - \frac{x ^{2}}{2} + C

Solution steps

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

Expand

x (x^{2} - 1)^{3} : x^{7} - 3 x^{5} + 3 x^{3} - x

= \int x^{7} - 3 x^{5} + 3 x^{3} - x 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 x^{7} d x - \int 3 x^{5} d x + \int 3 x^{3} d x - \int x d x

\int x^{7} d x = \frac{x ^{8}}{8}

\int 3 x^{5} d x = \frac{x ^{6}}{2}

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

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

= \frac{x ^{8}}{8} - \frac{x ^{6}}{2} + \frac{3 x ^{4}}{4} - \frac{x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{8}}{8} - \frac{x ^{6}}{2} + \frac{3 x ^{4}}{4} - \frac{x ^{2}}{2} + C

d er i v a t i v e f (x) = 2 - 3 x

y^{^{'}} - 3 y = x y^{- 4}

d er i v a t i v e T 4 \frac{7}{9}

s l o p e in t erce pt (3.1) (4.2)

\frac{d}{d x} (\frac{e ^{x}}{6 x ^{2} + 5})

What is the integral of x(x^2-1)^3 ?
The integral of x(x^2-1)^3 is (x^8)/8-(x^6)/2+(3x^4)/4-(x^2)/2+C