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

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

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

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

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

Solution steps

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

Expand

(x + 2)^{2} (1 - x)^{5} : - x^{7} + x^{6} + 6 x^{5} - 10 x^{4} - 5 x^{3} + 21 x^{2} - 16 x + 4

= \int - x^{7} + x^{6} + 6 x^{5} - 10 x^{4} - 5 x^{3} + 21 x^{2} - 16 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^{7} d x + \int x^{6} d x + \int 6 x^{5} d x - \int 10 x^{4} d x - \int 5 x^{3} d x + \int 21 x^{2} d x - \int 16 x d x + \int 4 d x

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

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

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

\int 10 x^{4} d x = 2 x^{5}

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

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

\int 16 x d x = 8 x^{2}

\int 4 d x = 4 x

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

Add a constant to the solution

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

y^{^{''}} - 2 y^{^{'}} - 3 y = t

\int \frac{1}{( v ^{2} + v ) ( v - 1 ) ^{2}} d v

\int 9 z^{3} e^{z} d z

f (x) = \frac{x ^{2} - 1}{( x ^{2} + x + 1 )}

\int ln (5)

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