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

The integral of 5(x^5-10x)^2(x^4-2) is 5((x^{15}}{15}-2x^{11}+20x^7-\frac{200x^3)/3)+C

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

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

5 (\frac{x ^{15}}{15} - 2 x^{11} + 20 x^{7} - \frac{200 x ^{3}}{3}) + C

Solution steps

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

Take the constant out:

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

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

Expand

(x^{5} - 10 x)^{2} (x^{4} - 2) : x^{14} - 22 x^{10} + 140 x^{6} - 200 x^{2}

= 5 \cdot \int x^{14} - 22 x^{10} + 140 x^{6} - 200 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 x^{14} d x - \int 22 x^{10} d x + \int 140 x^{6} d x - \int 200 x^{2} d x)

\int x^{14} d x = \frac{x ^{15}}{15}

\int 22 x^{10} d x = 2 x^{11}

\int 140 x^{6} d x = 20 x^{7}

\int 200 x^{2} d x = \frac{200 x ^{3}}{3}

= 5 (\frac{x ^{15}}{15} - 2 x^{11} + 20 x^{7} - \frac{200 x ^{3}}{3})

Add a constant to the solution

= 5 (\frac{x ^{15}}{15} - 2 x^{11} + 20 x^{7} - \frac{200 x ^{3}}{3}) + C

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f (x) = sin (4 x)

cot^{5} (x)

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What is the integral of 5(x^5-10x)^2(x^4-2) ?
The integral of 5(x^5-10x)^2(x^4-2) is 5((x^{15}}{15}-2x^{11}+20x^7-\frac{200x^3)/3)+C