What is the integral of x(x^2-1)^3 ?

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

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integral of x(x^2-1)^3

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

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

Solution steps

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

Expand

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

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

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

\int 3 x^{5} d x = \frac{x ^{6}}{2}

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

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

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

Add a constant to the solution

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

3 x^{2} - 26 x + 8

f (x) = 2 - 3 x

y^{^{'}} - 3 y = x y^{- 4}

\int tan^{4} (x) sec^{2} (x) d x

f (x) = x^{2} - 6 x, at x = - 2

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