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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

\int 1 d x = x

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

Add a constant to the solution

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

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