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

The integral of (3x(x^2-1)^3)/4 is 3/32 (x^8-4x^6+6x^4-4x^2)+C

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integral of (3x(x^2-1)^3)/4

\int \frac{3 x ( x ^{2} - 1 ) ^{3}}{4} d x

\frac{3}{32} (x^{8} - 4 x^{6} + 6 x^{4} - 4 x^{2}) + C

Solution steps

\int \frac{3 x ( x ^{2} - 1 ) ^{3}}{4} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{3}{4} \cdot \int x (x^{2} - 1)^{3} d x

Expand

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

= \frac{3}{4} \cdot \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

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

Simplify

\frac{3}{4} (\frac{x ^{8}}{8} - \frac{x ^{6}}{2} + \frac{3 x ^{4}}{4} - \frac{x ^{2}}{2}) : \frac{3}{32} (x^{8} - 4 x^{6} + 6 x^{4} - 4 x^{2})

= \frac{3}{32} (x^{8} - 4 x^{6} + 6 x^{4} - 4 x^{2})

Add a constant to the solution

= \frac{3}{32} (x^{8} - 4 x^{6} + 6 x^{4} - 4 x^{2}) + C

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What is the integral of (3x(x^2-1)^3)/4 ?
The integral of (3x(x^2-1)^3)/4 is 3/32 (x^8-4x^6+6x^4-4x^2)+C