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

The integral of (x-1)^7(x+1) is (x^9)/9-(3x^8)/4+2x^7-(7x^6)/3+(7x^4)/2-(14x^3)/3+3x^2-x+C

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integral of (x-1)^7(x+1)

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

\frac{x ^{9}}{9} - \frac{3 x ^{8}}{4} + 2 x^{7} - \frac{7 x ^{6}}{3} + \frac{7 x ^{4}}{2} - \frac{14 x ^{3}}{3} + 3 x^{2} - x + C

Solution steps

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

Expand

(x - 1)^{7} (x + 1) : x^{8} - 6 x^{7} + 14 x^{6} - 14 x^{5} + 14 x^{3} - 14 x^{2} + 6 x - 1

= \int x^{8} - 6 x^{7} + 14 x^{6} - 14 x^{5} + 14 x^{3} - 14 x^{2} + 6 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^{8} d x - \int 6 x^{7} d x + \int 14 x^{6} d x - \int 14 x^{5} d x + \int 14 x^{3} d x - \int 14 x^{2} d x + \int 6 x d x - \int 1 d x

\int x^{8} d x = \frac{x ^{9}}{9}

\int 6 x^{7} d x = \frac{3 x ^{8}}{4}

\int 14 x^{6} d x = 2 x^{7}

\int 14 x^{5} d x = \frac{7 x ^{6}}{3}

\int 14 x^{3} d x = \frac{7 x ^{4}}{2}

\int 14 x^{2} d x = \frac{14 x ^{3}}{3}

\int 6 x d x = 3 x^{2}

\int 1 d x = x

= \frac{x ^{9}}{9} - \frac{3 x ^{8}}{4} + 2 x^{7} - \frac{7 x ^{6}}{3} + \frac{7 x ^{4}}{2} - \frac{14 x ^{3}}{3} + 3 x^{2} - x

Add a constant to the solution

= \frac{x ^{9}}{9} - \frac{3 x ^{8}}{4} + 2 x^{7} - \frac{7 x ^{6}}{3} + \frac{7 x ^{4}}{2} - \frac{14 x ^{3}}{3} + 3 x^{2} - x + C

What is the integral of (x-1)^7(x+1) ?
The integral of (x-1)^7(x+1) is (x^9)/9-(3x^8)/4+2x^7-(7x^6)/3+(7x^4)/2-(14x^3)/3+3x^2-x+C