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

The integral of-x^2(x+1)^3(x-1) is -(x^7)/7-(x^6)/3+(x^4)/2+(x^3)/3+C

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

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

- \frac{x ^{7}}{7} - \frac{x ^{6}}{3} + \frac{x ^{4}}{2} + \frac{x ^{3}}{3} + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Expand

x^{2} (x + 1)^{3} (x - 1) : x^{6} + 2 x^{5} - 2 x^{3} - x^{2}

= - \int x^{6} + 2 x^{5} - 2 x^{3} - x^{2} 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^{6} d x + \int 2 x^{5} d x - \int 2 x^{3} d x - \int x^{2} d x)

\int x^{6} d x = \frac{x ^{7}}{7}

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

\int 2 x^{3} d x = \frac{x ^{4}}{2}

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

= - (\frac{x ^{7}}{7} + \frac{x ^{6}}{3} - \frac{x ^{4}}{2} - \frac{x ^{3}}{3})

Simplify

= - \frac{x ^{7}}{7} - \frac{x ^{6}}{3} + \frac{x ^{4}}{2} + \frac{x ^{3}}{3}

Add a constant to the solution

= - \frac{x ^{7}}{7} - \frac{x ^{6}}{3} + \frac{x ^{4}}{2} + \frac{x ^{3}}{3} + C

\frac{\partial}{\partial y} (x^{2} + 2 y^{2})

\frac{d}{d x} (arctan (\frac{1 - x}{1 + x}))

\int_{0}^{3} (- x^{2} + x - 1) d x

\int_{1}^{16} \frac{1}{x} d x

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

What is the integral of-x^2(x+1)^3(x-1) ?
The integral of-x^2(x+1)^3(x-1) is -(x^7)/7-(x^6)/3+(x^4)/2+(x^3)/3+C