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

The integral of (x-1)(x+3)^3 is (x^5)/5+2x^4+6x^3-27x+C

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

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

\frac{x ^{5}}{5} + 2 x^{4} + 6 x^{3} - 27 x + C

Solution steps

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

Expand

(x - 1) (x + 3)^{3} : x^{4} + 8 x^{3} + 18 x^{2} - 27

= \int x^{4} + 8 x^{3} + 18 x^{2} - 27 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^{4} d x + \int 8 x^{3} d x + \int 18 x^{2} d x - \int 27 d x

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

\int 8 x^{3} d x = 2 x^{4}

\int 18 x^{2} d x = 6 x^{3}

\int 27 d x = 27 x

= \frac{x ^{5}}{5} + 2 x^{4} + 6 x^{3} - 27 x

Add a constant to the solution

= \frac{x ^{5}}{5} + 2 x^{4} + 6 x^{3} - 27 x + C

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What is the integral of (x-1)(x+3)^3 ?
The integral of (x-1)(x+3)^3 is (x^5)/5+2x^4+6x^3-27x+C