What is the integral of (y^2+4y)^2 ?

The integral of (y^2+4y)^2 is (y^5)/5+2y^4+(16y^3)/3+C

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integral of (y^2+4y)^2

\int (y^{2} + 4 y)^{2} d y

\frac{y ^{5}}{5} + 2 y^{4} + \frac{16 y ^{3}}{3} + C

Solution steps

\int (y^{2} + 4 y)^{2} d y

Expand

(y^{2} + 4 y)^{2} : y^{4} + 8 y^{3} + 16 y^{2}

= \int y^{4} + 8 y^{3} + 16 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

= \int y^{4} d y + \int 8 y^{3} d y + \int 16 y^{2} d y

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

\int 8 y^{3} d y = 2 y^{4}

\int 16 y^{2} d y = \frac{16 y ^{3}}{3}

= \frac{y ^{5}}{5} + 2 y^{4} + \frac{16 y ^{3}}{3}

Add a constant to the solution

= \frac{y ^{5}}{5} + 2 y^{4} + \frac{16 y ^{3}}{3} + C

5 x^{3} + 2

y^{^{'}} = y x e^{x + 2} - y

x y^{^{''}} + 2 y^{^{'}} = 20 x^{2}

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

What is the integral of (y^2+4y)^2 ?
The integral of (y^2+4y)^2 is (y^5)/5+2y^4+(16y^3)/3+C