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

The integral of (x^2-1)^2(3x^2) is (3x^7)/7-(6x^5)/5+x^3+C

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

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

\frac{3 x ^{7}}{7} - \frac{6 x ^{5}}{5} + x^{3} + C

Solution steps

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

Take the constant out:

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

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

Expand

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

= 3 \cdot \int x^{6} - 2 x^{4} + 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

= 3 (\int x^{6} d x - \int 2 x^{4} d x + \int x^{2} d x)

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

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

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

= 3 (\frac{x ^{7}}{7} - \frac{2 x ^{5}}{5} + \frac{x ^{3}}{3})

Simplify

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

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

Add a constant to the solution

= \frac{3 x ^{7}}{7} - \frac{6 x ^{5}}{5} + x^{3} + C

\frac{d x}{d y} = e^{3 x + 2 y}

\frac{d}{d x} (\frac{x}{x + 16})

x \to 3 lim (\frac{0}{x - 3})

n = 0 \sum \infty 1 1^{n}

\frac{d}{d x} (ln (x^{3} + x))

What is the integral of (x^2-1)^2(3x^2) ?
The integral of (x^2-1)^2(3x^2) is (3x^7)/7-(6x^5)/5+x^3+C