What is the integral of y(y+1)^2 ?

The integral of y(y+1)^2 is (y^4)/4+(2y^3)/3+(y^2)/2+C

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integral of y(y+1)^2

\int y (y + 1)^{2} d y

\frac{y ^{4}}{4} + \frac{2 y ^{3}}{3} + \frac{y ^{2}}{2} + C

Solution steps

\int y (y + 1)^{2} d y

Expand

y (y + 1)^{2} : y^{3} + 2 y^{2} + y

= \int y^{3} + 2 y^{2} + y 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

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

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

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

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

= \frac{y ^{4}}{4} + \frac{2 y ^{3}}{3} + \frac{y ^{2}}{2}

Add a constant to the solution

= \frac{y ^{4}}{4} + \frac{2 y ^{3}}{3} + \frac{y ^{2}}{2} + C

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What is the integral of y(y+1)^2 ?
The integral of y(y+1)^2 is (y^4)/4+(2y^3)/3+(y^2)/2+C