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

The integral of ((x-1)^2-(x+1)^2)(x^3+1) is -4((x^5)/5+(x^2)/2)+C

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

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

- 4 (\frac{x ^{5}}{5} + \frac{x ^{2}}{2}) + C

Solution steps

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

Simplify

((x - 1)^{2} - (x + 1)^{2}) (x^{3} + 1) : - 4 x (x^{3} + 1)

= \int - 4 x (x^{3} + 1) d x

Take the constant out:

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

= - 4 \cdot \int x (x^{3} + 1) d x

Expand

x (x^{3} + 1) : x^{4} + x

= - 4 \cdot \int x^{4} + 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

= - 4 (\int x^{4} d x + \int x d x)

\int x^{4} d x = \frac{x ^{5}}{5}

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

= - 4 (\frac{x ^{5}}{5} + \frac{x ^{2}}{2})

Add a constant to the solution

= - 4 (\frac{x ^{5}}{5} + \frac{x ^{2}}{2}) + C

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

\frac{d}{d x} (x^{2} arcsinh (9 x))

\int \frac{1}{x ^{5}} - \frac{3}{x ^{2}} d x

x y^{^{'}} + 2 y = x^{2} - x + 1

What is the integral of ((x-1)^2-(x+1)^2)(x^3+1) ?
The integral of ((x-1)^2-(x+1)^2)(x^3+1) is -4((x^5)/5+(x^2)/2)+C