What is the integral of [(x^2-1)(x^3+7x)] ?

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

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integral of [(x^2-1)(x^3+7x)]

\int [(x^{2} - 1) (x^{3} + 7 x)] d x

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

Solution steps

\int (x^{2} - 1) (x^{3} + 7 x) d x

Expand

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

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

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

\int 6 x^{3} d x = \frac{3 x ^{4}}{2}

\int 7 x d x = \frac{7 x ^{2}}{2}

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

Add a constant to the solution

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

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