What is the integral of (x-1)^2(x-2)(x-4) ?

The integral of (x-1)^2(x-2)(x-4) is (x^5)/5-2x^4+7x^3-11x^2+8x+C

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integral of (x-1)^2(x-2)(x-4)

\int (x - 1)^{2} (x - 2) (x - 4) d x

\frac{x ^{5}}{5} - 2 x^{4} + 7 x^{3} - 11 x^{2} + 8 x + C

Solution steps

\int (x - 1)^{2} (x - 2) (x - 4) d x

Expand

(x - 1)^{2} (x - 2) (x - 4) : x^{4} - 8 x^{3} + 21 x^{2} - 22 x + 8

= \int x^{4} - 8 x^{3} + 21 x^{2} - 22 x + 8 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 21 x^{2} d x - \int 22 x d x + \int 8 d x

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

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

\int 21 x^{2} d x = 7 x^{3}

\int 22 x d x = 11 x^{2}

\int 8 d x = 8 x

= \frac{x ^{5}}{5} - 2 x^{4} + 7 x^{3} - 11 x^{2} + 8 x

Add a constant to the solution

= \frac{x ^{5}}{5} - 2 x^{4} + 7 x^{3} - 11 x^{2} + 8 x + C

x^{2} - 7 x

\frac{\partial}{\partial x} (\frac{7 x}{4 + x ^{2} + y ^{2}})

(lo g_{10} (1 + x))^{^{'}}

- 8 x

\int sin (x) - sin^{3} (x) d x

What is the integral of (x-1)^2(x-2)(x-4) ?
The integral of (x-1)^2(x-2)(x-4) is (x^5)/5-2x^4+7x^3-11x^2+8x+C