What is the integral of (3x^3+6x)^3(3x^2+2) ?

The integral of (3x^3+6x)^3(3x^2+2) is (27x^{12})/4+54x^{10}+162x^8+216x^6+108x^4+C

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integral of (3x^3+6x)^3(3x^2+2)

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

\frac{27 x ^{12}}{4} + 54 x^{10} + 162 x^{8} + 216 x^{6} + 108 x^{4} + C

Solution steps

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

Expand

(3 x^{3} + 6 x)^{3} (3 x^{2} + 2) : 81 x^{11} + 540 x^{9} + 1296 x^{7} + 1296 x^{5} + 432 x^{3}

= \int 81 x^{11} + 540 x^{9} + 1296 x^{7} + 1296 x^{5} + 432 x^{3} 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 81 x^{11} d x + \int 540 x^{9} d x + \int 1296 x^{7} d x + \int 1296 x^{5} d x + \int 432 x^{3} d x

\int 81 x^{11} d x = \frac{27 x ^{12}}{4}

\int 540 x^{9} d x = 54 x^{10}

\int 1296 x^{7} d x = 162 x^{8}

\int 1296 x^{5} d x = 216 x^{6}

\int 432 x^{3} d x = 108 x^{4}

= \frac{27 x ^{12}}{4} + 54 x^{10} + 162 x^{8} + 216 x^{6} + 108 x^{4}

Add a constant to the solution

= \frac{27 x ^{12}}{4} + 54 x^{10} + 162 x^{8} + 216 x^{6} + 108 x^{4} + C

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What is the integral of (3x^3+6x)^3(3x^2+2) ?
The integral of (3x^3+6x)^3(3x^2+2) is (27x^{12})/4+54x^{10}+162x^8+216x^6+108x^4+C