What is the integral of-5x^2(3x-8)^4 ?

The integral of-5x^2(3x-8)^4 is -(405x^7)/7+720x^6-3456x^5+7680x^4-(20480x^3)/3+C

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integral of-5x^2(3x-8)^4

\int - 5 x^{2} (3 x - 8)^{4} d x

- \frac{405 x ^{7}}{7} + 720 x^{6} - 3456 x^{5} + 7680 x^{4} - \frac{20480 x ^{3}}{3} + C

Solution steps

\int - 5 x^{2} (3 x - 8)^{4} d x

Take the constant out:

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

= - 5 \cdot \int x^{2} (3 x - 8)^{4} d x

Factor

3 x - 8 : - (- 3 x + 8)

= - 5 \cdot \int x^{2} (- (- 3 x + 8))^{4} d x

(- (- 3 x + 8))^{4} = (- 3 x + 8)^{4}

= - 5 \cdot \int x^{2} (- 3 x + 8)^{4} d x

Expand

x^{2} (- 3 x + 8)^{4} : 81 x^{6} - 864 x^{5} + 3456 x^{4} - 6144 x^{3} + 4096 x^{2}

= - 5 \cdot \int 81 x^{6} - 864 x^{5} + 3456 x^{4} - 6144 x^{3} + 4096 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

= - 5 (\int 81 x^{6} d x - \int 864 x^{5} d x + \int 3456 x^{4} d x - \int 6144 x^{3} d x + \int 4096 x^{2} d x)

\int 81 x^{6} d x = \frac{81 x ^{7}}{7}

\int 864 x^{5} d x = 144 x^{6}

\int 3456 x^{4} d x = \frac{3456 x ^{5}}{5}

\int 6144 x^{3} d x = 1536 x^{4}

\int 4096 x^{2} d x = \frac{4096 x ^{3}}{3}

= - 5 (\frac{81 x ^{7}}{7} - 144 x^{6} + \frac{3456 x ^{5}}{5} - 1536 x^{4} + \frac{4096 x ^{3}}{3})

Simplify

- 5 (\frac{81 x ^{7}}{7} - 144 x^{6} + \frac{3456 x ^{5}}{5} - 1536 x^{4} + \frac{4096 x ^{3}}{3}) : - \frac{405 x ^{7}}{7} + 720 x^{6} - 3456 x^{5} + 7680 x^{4} - \frac{20480 x ^{3}}{3}

= - \frac{405 x ^{7}}{7} + 720 x^{6} - 3456 x^{5} + 7680 x^{4} - \frac{20480 x ^{3}}{3}

Add a constant to the solution

= - \frac{405 x ^{7}}{7} + 720 x^{6} - 3456 x^{5} + 7680 x^{4} - \frac{20480 x ^{3}}{3} + C

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What is the integral of-5x^2(3x-8)^4 ?
The integral of-5x^2(3x-8)^4 is -(405x^7)/7+720x^6-3456x^5+7680x^4-(20480x^3)/3+C