What is the integral of (x-1)(x-4) ?

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

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integral of (x-1)(x-4)

\int (x - 1) (x - 4) d x

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

Solution steps

\int (x - 1) (x - 4) d x

Expand

(x - 1) (x - 4) : x^{2} - 5 x + 4

= \int x^{2} - 5 x + 4 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^{2} d x - \int 5 x d x + \int 4 d x

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

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

\int 4 d x = 4 x

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

Add a constant to the solution

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

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\frac{d}{d x} (x + \frac{4}{x + 3})

\frac{d}{d x} ((tan (x))^{x})

\frac{d y}{d x} = \frac{cos ( 6 x )}{e ^{6 y}}

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