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

The integral of (x+3)^2(2-x)^5 is -(x^8)/8+(4x^7)/7+(11x^6)/6-14x^5+10x^4+(272x^3)/3-264x^2+288x+C

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

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

- \frac{x ^{8}}{8} + \frac{4 x ^{7}}{7} + \frac{11 x ^{6}}{6} - 14 x^{5} + 10 x^{4} + \frac{272 x ^{3}}{3} - 264 x^{2} + 288 x + C

Solution steps

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

Expand

(x + 3)^{2} (2 - x)^{5} : - x^{7} + 4 x^{6} + 11 x^{5} - 70 x^{4} + 40 x^{3} + 272 x^{2} - 528 x + 288

= \int - x^{7} + 4 x^{6} + 11 x^{5} - 70 x^{4} + 40 x^{3} + 272 x^{2} - 528 x + 288 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^{7} d x + \int 4 x^{6} d x + \int 11 x^{5} d x - \int 70 x^{4} d x + \int 40 x^{3} d x + \int 272 x^{2} d x - \int 528 x d x + \int 288 d x

\int x^{7} d x = \frac{x ^{8}}{8}

\int 4 x^{6} d x = \frac{4 x ^{7}}{7}

\int 11 x^{5} d x = \frac{11 x ^{6}}{6}

\int 70 x^{4} d x = 14 x^{5}

\int 40 x^{3} d x = 10 x^{4}

\int 272 x^{2} d x = \frac{272 x ^{3}}{3}

\int 528 x d x = 264 x^{2}

\int 288 d x = 288 x

= - \frac{x ^{8}}{8} + \frac{4 x ^{7}}{7} + \frac{11 x ^{6}}{6} - 14 x^{5} + 10 x^{4} + \frac{272 x ^{3}}{3} - 264 x^{2} + 288 x

Add a constant to the solution

= - \frac{x ^{8}}{8} + \frac{4 x ^{7}}{7} + \frac{11 x ^{6}}{6} - 14 x^{5} + 10 x^{4} + \frac{272 x ^{3}}{3} - 264 x^{2} + 288 x + C

y^{^{''}} + 6 y + 9 = 0, y (0) = - 3, y^{^{'}} (0) = - 1

\int 4 x^{4} d x

\frac{\partial}{\partial z} (x e^{y} + y e^{z} - z e^{x})

\frac{d}{d x} (8 sin^{6} (x))

n = 3 \sum \infty \frac{1}{2 ^{n}}

What is the integral of (x+3)^2(2-x)^5 ?
The integral of (x+3)^2(2-x)^5 is -(x^8)/8+(4x^7)/7+(11x^6)/6-14x^5+10x^4+(272x^3)/3-264x^2+288x+C