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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x(-1+x^2)

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

- \frac{x ^{2}}{2} + \frac{x ^{4}}{4} + C

Solution steps

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

Expand

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

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

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

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

= - \frac{x ^{2}}{2} + \frac{x ^{4}}{4}

Add a constant to the solution

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

x \to 0 lim (\frac{4 x - sin ( x )}{x})

\int e^{y} y^{2} d y

\frac{\partial}{\partial x} (9 x e^{x + y})

y^{^{''}} + 4 y^{^{'}} + 5 y = - 5 x + 5 e^{- x}

\int \frac{3}{2} e^{2 x} d x

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