What is the integral of 6x(4x^2-9)^2 ?

The integral of 6x(4x^2-9)^2 is 6((8x^6)/3-18x^4+(81x^2)/2)+C

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integral of 6x(4x^2-9)^2

\int 6 x (4 x^{2} - 9)^{2} d x

6 (\frac{8 x ^{6}}{3} - 18 x^{4} + \frac{81 x ^{2}}{2}) + C

Solution steps

\int 6 x (4 x^{2} - 9)^{2} d x

Take the constant out:

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

= 6 \cdot \int x (4 x^{2} - 9)^{2} d x

Expand

x (4 x^{2} - 9)^{2} : 16 x^{5} - 72 x^{3} + 81 x

= 6 \cdot \int 16 x^{5} - 72 x^{3} + 81 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

= 6 (\int 16 x^{5} d x - \int 72 x^{3} d x + \int 81 x d x)

\int 16 x^{5} d x = \frac{8 x ^{6}}{3}

\int 72 x^{3} d x = 18 x^{4}

\int 81 x d x = \frac{81 x ^{2}}{2}

= 6 (\frac{8 x ^{6}}{3} - 18 x^{4} + \frac{81 x ^{2}}{2})

Add a constant to the solution

= 6 (\frac{8 x ^{6}}{3} - 18 x^{4} + \frac{81 x ^{2}}{2}) + C

f (x) = 3 x - 0.6 x ln (x)

x \to 2 lim (x^{2} - 8 x - 2)

\frac{\partial}{\partial x} (3 x^{2} + 2 x y - 7 y^{2})

f (x) = 3 - 5 x^{2}

f^{^{'}} (x) = x

What is the integral of 6x(4x^2-9)^2 ?
The integral of 6x(4x^2-9)^2 is 6((8x^6)/3-18x^4+(81x^2)/2)+C