What is the integral of (x^4-1)x^2 ?

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

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integral of (x^4-1)x^2

\int (x^{4} - 1) x^{2} d x

\frac{x ^{7}}{7} - \frac{x ^{3}}{3} + C

Solution steps

\int (x^{4} - 1) x^{2} d x

Expand

(x^{4} - 1) x^{2} : x^{6} - x^{2}

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

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

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

= \frac{x ^{7}}{7} - \frac{x ^{3}}{3}

Add a constant to the solution

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

\frac{d y}{d x} = y (2 - 3 y)

f (x) = ln (x^{2} + 2)

\int 2 sin (x) - 3 cos (x) d x

f (x) = \frac{x ^{2}}{x + 2}, at x = 3

\int (6 x^{6} - 2)^{2} (36 x) d x

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