What is the integral of 12x^2(4x^3-3)^5 ?

The integral of 12x^2(4x^3-3)^5 is 12((512x^{18})/9-256x^{15}+480x^{12}-480x^9+270x^6-81x^3)+C

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integral of 12x^2(4x^3-3)^5

\int 12 x^{2} (4 x^{3} - 3)^{5} d x

12 (\frac{512 x ^{18}}{9} - 256 x^{15} + 480 x^{12} - 480 x^{9} + 270 x^{6} - 81 x^{3}) + C

Solution steps

\int 12 x^{2} (4 x^{3} - 3)^{5} d x

Take the constant out:

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

= 12 \cdot \int x^{2} (4 x^{3} - 3)^{5} d x

Expand

x^{2} (4 x^{3} - 3)^{5} : 1024 x^{17} - 3840 x^{14} + 5760 x^{11} - 4320 x^{8} + 1620 x^{5} - 243 x^{2}

= 12 \cdot \int 1024 x^{17} - 3840 x^{14} + 5760 x^{11} - 4320 x^{8} + 1620 x^{5} - 243 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

= 12 (\int 1024 x^{17} d x - \int 3840 x^{14} d x + \int 5760 x^{11} d x - \int 4320 x^{8} d x + \int 1620 x^{5} d x - \int 243 x^{2} d x)

\int 1024 x^{17} d x = \frac{512 x ^{18}}{9}

\int 3840 x^{14} d x = 256 x^{15}

\int 5760 x^{11} d x = 480 x^{12}

\int 4320 x^{8} d x = 480 x^{9}

\int 1620 x^{5} d x = 270 x^{6}

\int 243 x^{2} d x = 81 x^{3}

= 12 (\frac{512 x ^{18}}{9} - 256 x^{15} + 480 x^{12} - 480 x^{9} + 270 x^{6} - 81 x^{3})

Add a constant to the solution

= 12 (\frac{512 x ^{18}}{9} - 256 x^{15} + 480 x^{12} - 480 x^{9} + 270 x^{6} - 81 x^{3}) + C

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What is the integral of 12x^2(4x^3-3)^5 ?
The integral of 12x^2(4x^3-3)^5 is 12((512x^{18})/9-256x^{15}+480x^{12}-480x^9+270x^6-81x^3)+C