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

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

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

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

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

Solution steps

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

Expand

(x^{2} + 2)^{2} (x - 2) : x^{5} - 2 x^{4} + 4 x^{3} - 8 x^{2} + 4 x - 8

= \int x^{5} - 2 x^{4} + 4 x^{3} - 8 x^{2} + 4 x - 8 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^{5} d x - \int 2 x^{4} d x + \int 4 x^{3} d x - \int 8 x^{2} d x + \int 4 x d x - \int 8 d x

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

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

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

\int 8 x^{2} d x = \frac{8 x ^{3}}{3}

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

\int 8 d x = 8 x

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

Add a constant to the solution

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

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