What is the integral of 4x^2(x-1)^5 ?

The integral of 4x^2(x-1)^5 is (x^8)/2-(20x^7)/7+(20x^6)/3-8x^5+5x^4-(4x^3)/3+C

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integral of 4x^2(x-1)^5

\int 4 x^{2} (x - 1)^{5} d x

\frac{x ^{8}}{2} - \frac{20 x ^{7}}{7} + \frac{20 x ^{6}}{3} - 8 x^{5} + 5 x^{4} - \frac{4 x ^{3}}{3} + C

Solution steps

\int 4 x^{2} (x - 1)^{5} d x

Take the constant out:

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

= 4 \cdot \int x^{2} (x - 1)^{5} d x

Expand

x^{2} (x - 1)^{5} : x^{7} - 5 x^{6} + 10 x^{5} - 10 x^{4} + 5 x^{3} - x^{2}

= 4 \cdot \int x^{7} - 5 x^{6} + 10 x^{5} - 10 x^{4} + 5 x^{3} - 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

= 4 (\int x^{7} d x - \int 5 x^{6} d x + \int 10 x^{5} d x - \int 10 x^{4} d x + \int 5 x^{3} d x - \int x^{2} d x)

\int x^{7} d x = \frac{x ^{8}}{8}

\int 5 x^{6} d x = \frac{5 x ^{7}}{7}

\int 10 x^{5} d x = \frac{5 x ^{6}}{3}

\int 10 x^{4} d x = 2 x^{5}

\int 5 x^{3} d x = \frac{5 x ^{4}}{4}

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

= 4 (\frac{x ^{8}}{8} - \frac{5 x ^{7}}{7} + \frac{5 x ^{6}}{3} - 2 x^{5} + \frac{5 x ^{4}}{4} - \frac{x ^{3}}{3})

Simplify

4 (\frac{x ^{8}}{8} - \frac{5 x ^{7}}{7} + \frac{5 x ^{6}}{3} - 2 x^{5} + \frac{5 x ^{4}}{4} - \frac{x ^{3}}{3}) : \frac{x ^{8}}{2} - \frac{20 x ^{7}}{7} + \frac{20 x ^{6}}{3} - 8 x^{5} + 5 x^{4} - \frac{4 x ^{3}}{3}

= \frac{x ^{8}}{2} - \frac{20 x ^{7}}{7} + \frac{20 x ^{6}}{3} - 8 x^{5} + 5 x^{4} - \frac{4 x ^{3}}{3}

Add a constant to the solution

= \frac{x ^{8}}{2} - \frac{20 x ^{7}}{7} + \frac{20 x ^{6}}{3} - 8 x^{5} + 5 x^{4} - \frac{4 x ^{3}}{3} + C

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What is the integral of 4x^2(x-1)^5 ?
The integral of 4x^2(x-1)^5 is (x^8)/2-(20x^7)/7+(20x^6)/3-8x^5+5x^4-(4x^3)/3+C