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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Expand

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

= 4 \cdot \int y^{7} - 5 y^{6} + 10 y^{5} - 10 y^{4} + 5 y^{3} - y^{2} d y

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

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

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

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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