What is the integral of 9x^2(3x^3-4)^5 ?

The integral of 9x^2(3x^3-4)^5 is 9((27x^{18})/2-108x^{15}+360x^{12}-640x^9+640x^6-(1024x^3)/3)+C

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integral of 9x^2(3x^3-4)^5

\int 9 x^{2} (3 x^{3} - 4)^{5} d x

9 (\frac{27 x ^{18}}{2} - 108 x^{15} + 360 x^{12} - 640 x^{9} + 640 x^{6} - \frac{1024 x ^{3}}{3}) + C

Solution steps

\int 9 x^{2} (3 x^{3} - 4)^{5} d x

Take the constant out:

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

= 9 \cdot \int x^{2} (3 x^{3} - 4)^{5} d x

Expand

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

= 9 \cdot \int 243 x^{17} - 1620 x^{14} + 4320 x^{11} - 5760 x^{8} + 3840 x^{5} - 1024 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

= 9 (\int 243 x^{17} d x - \int 1620 x^{14} d x + \int 4320 x^{11} d x - \int 5760 x^{8} d x + \int 3840 x^{5} d x - \int 1024 x^{2} d x)

\int 243 x^{17} d x = \frac{27 x ^{18}}{2}

\int 1620 x^{14} d x = 108 x^{15}

\int 4320 x^{11} d x = 360 x^{12}

\int 5760 x^{8} d x = 640 x^{9}

\int 3840 x^{5} d x = 640 x^{6}

\int 1024 x^{2} d x = \frac{1024 x ^{3}}{3}

= 9 (\frac{27 x ^{18}}{2} - 108 x^{15} + 360 x^{12} - 640 x^{9} + 640 x^{6} - \frac{1024 x ^{3}}{3})

Add a constant to the solution

= 9 (\frac{27 x ^{18}}{2} - 108 x^{15} + 360 x^{12} - 640 x^{9} + 640 x^{6} - \frac{1024 x ^{3}}{3}) + C

What is the integral of 9x^2(3x^3-4)^5 ?
The integral of 9x^2(3x^3-4)^5 is 9((27x^{18})/2-108x^{15}+360x^{12}-640x^9+640x^6-(1024x^3)/3)+C