What is the integral of (6x^6-2)^2(36x) ?

The integral of (6x^6-2)^2(36x) is 36((18x^{14})/7-3x^8+2x^2)+C

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integral of (6x^6-2)^2(36x)

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

36 (\frac{18 x ^{14}}{7} - 3 x^{8} + 2 x^{2}) + C

Solution steps

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

Take the constant out:

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

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

Expand

(6 x^{6} - 2)^{2} x : 36 x^{13} - 24 x^{7} + 4 x

= 36 \cdot \int 36 x^{13} - 24 x^{7} + 4 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

= 36 (\int 36 x^{13} d x - \int 24 x^{7} d x + \int 4 x d x)

\int 36 x^{13} d x = \frac{18 x ^{14}}{7}

\int 24 x^{7} d x = 3 x^{8}

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

= 36 (\frac{18 x ^{14}}{7} - 3 x^{8} + 2 x^{2})

Add a constant to the solution

= 36 (\frac{18 x ^{14}}{7} - 3 x^{8} + 2 x^{2}) + C

\int x^{3} x^{2} + 6 d x

\frac{d y}{d x} = \frac{2 x - 1}{y ^{2}}

\frac{\partial}{\partial x} (x y (1 - 7 x - 6 y))

- \frac{( x ^{2} )}{y ^{2}}

\int \frac{( 1 + x ) ^{2}}{x ^{2} + 1} d x

What is the integral of (6x^6-2)^2(36x) ?
The integral of (6x^6-2)^2(36x) is 36((18x^{14})/7-3x^8+2x^2)+C