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

The integral of (x+1)^2(1-x)^5 is -(x^8)/8+(3x^7)/7-(x^6)/6-x^5+(5x^4)/4+(x^3)/3-(3x^2)/2+x+C

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

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

- \frac{x ^{8}}{8} + \frac{3 x ^{7}}{7} - \frac{x ^{6}}{6} - x^{5} + \frac{5 x ^{4}}{4} + \frac{x ^{3}}{3} - \frac{3 x ^{2}}{2} + x + C

Solution steps

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

Expand

(x + 1)^{2} (1 - x)^{5} : - x^{7} + 3 x^{6} - x^{5} - 5 x^{4} + 5 x^{3} + x^{2} - 3 x + 1

= \int - x^{7} + 3 x^{6} - x^{5} - 5 x^{4} + 5 x^{3} + x^{2} - 3 x + 1 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 3 x^{6} d x - \int x^{5} d x - \int 5 x^{4} d x + \int 5 x^{3} d x + \int x^{2} d x - \int 3 x d x + \int 1 d x

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

\int 3 x^{6} d x = \frac{3 x ^{7}}{7}

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

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

\int 5 x^{3} d x = \frac{5 x ^{4}}{4}

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

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

\int 1 d x = x

= - \frac{x ^{8}}{8} + \frac{3 x ^{7}}{7} - \frac{x ^{6}}{6} - x^{5} + \frac{5 x ^{4}}{4} + \frac{x ^{3}}{3} - \frac{3 x ^{2}}{2} + x

Add a constant to the solution

= - \frac{x ^{8}}{8} + \frac{3 x ^{7}}{7} - \frac{x ^{6}}{6} - x^{5} + \frac{5 x ^{4}}{4} + \frac{x ^{3}}{3} - \frac{3 x ^{2}}{2} + x + C

e^{c o s (3 x^{2} + x)}

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

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

f (x) = (2)^{x}, at x = 6

y = e^{6 x} cos (π x), (0, 1)

What is the integral of (x+1)^2(1-x)^5 ?
The integral of (x+1)^2(1-x)^5 is -(x^8)/8+(3x^7)/7-(x^6)/6-x^5+(5x^4)/4+(x^3)/3-(3x^2)/2+x+C