What is the integral of-7(x+3)^2(x-1)(x-5) ?

The integral of-7(x+3)^2(x-1)(x-5) is -7((x^5)/5-(22x^3)/3-12x^2+45x)+C

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integral of-7(x+3)^2(x-1)(x-5)

\int - 7 (x + 3)^{2} (x - 1) (x - 5) d x

- 7 (\frac{x ^{5}}{5} - \frac{22 x ^{3}}{3} - 12 x^{2} + 45 x) + C

Solution steps

\int - 7 (x + 3)^{2} (x - 1) (x - 5) d x

Take the constant out:

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

= - 7 \cdot \int (x + 3)^{2} (x - 1) (x - 5) d x

x + 3 = 3 (\frac{x}{3} + 1)

= - 7 \cdot \int (3 (\frac{x}{3} + 1))^{2} (x - 1) (x - 5) d x

Simplify

(3 (\frac{x}{3} + 1))^{2} (x - 1) (x - 5) : x^{4} - 22 x^{2} - 24 x + 45

= - 7 \cdot \int x^{4} - 22 x^{2} - 24 x + 45 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

= - 7 (\int x^{4} d x - \int 22 x^{2} d x - \int 24 x d x + \int 45 d x)

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

\int 22 x^{2} d x = \frac{22 x ^{3}}{3}

\int 24 x d x = 12 x^{2}

\int 45 d x = 45 x

= - 7 (\frac{x ^{5}}{5} - \frac{22 x ^{3}}{3} - 12 x^{2} + 45 x)

Add a constant to the solution

= - 7 (\frac{x ^{5}}{5} - \frac{22 x ^{3}}{3} - 12 x^{2} + 45 x) + C

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What is the integral of-7(x+3)^2(x-1)(x-5) ?
The integral of-7(x+3)^2(x-1)(x-5) is -7((x^5)/5-(22x^3)/3-12x^2+45x)+C