What is the integral of 10x^4(3-2x^5)^3 ?

The integral of 10x^4(3-2x^5)^3 is 10((27x^5)/5-(27x^{10})/5+(12x^{15})/5-(2x^{20})/5)+C

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integral of 10x^4(3-2x^5)^3

\int 10 x^{4} (3 - 2 x^{5})^{3} d x

10 (\frac{27 x ^{5}}{5} - \frac{27 x ^{10}}{5} + \frac{12 x ^{15}}{5} - \frac{2 x ^{20}}{5}) + C

Solution steps

\int 10 x^{4} (3 - 2 x^{5})^{3} d x

Take the constant out:

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

= 10 \cdot \int x^{4} (3 - 2 x^{5})^{3} d x

Expand

x^{4} (3 - 2 x^{5})^{3} : 27 x^{4} - 54 x^{9} + 36 x^{14} - 8 x^{19}

= 10 \cdot \int 27 x^{4} - 54 x^{9} + 36 x^{14} - 8 x^{19} 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

= 10 (\int 27 x^{4} d x - \int 54 x^{9} d x + \int 36 x^{14} d x - \int 8 x^{19} d x)

\int 27 x^{4} d x = \frac{27 x ^{5}}{5}

\int 54 x^{9} d x = \frac{27 x ^{10}}{5}

\int 36 x^{14} d x = \frac{12 x ^{15}}{5}

\int 8 x^{19} d x = \frac{2 x ^{20}}{5}

= 10 (\frac{27 x ^{5}}{5} - \frac{27 x ^{10}}{5} + \frac{12 x ^{15}}{5} - \frac{2 x ^{20}}{5})

Add a constant to the solution

= 10 (\frac{27 x ^{5}}{5} - \frac{27 x ^{10}}{5} + \frac{12 x ^{15}}{5} - \frac{2 x ^{20}}{5}) + C

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What is the integral of 10x^4(3-2x^5)^3 ?
The integral of 10x^4(3-2x^5)^3 is 10((27x^5)/5-(27x^{10})/5+(12x^{15})/5-(2x^{20})/5)+C