Solve ∫ left(y^2+4yright)^2 wrt y

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

Evaluate

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

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y^{2}+4y\right)^{2}.

$$\int \left(y^{2}\right)^{2}+8y^{2}y+16y^{2}\mathrm{d}y$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\int y^{4}+8y^{2}y+16y^{2}\mathrm{d}y$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Since \int y^{k}\mathrm{d}y=\frac{y^{k+1}}{k+1} for k\neq -1, replace \int y^{4}\mathrm{d}y with \frac{y^{5}}{5}.

$$\frac{y^{5}}{5}+8\int y^{3}\mathrm{d}y+16\int y^{2}\mathrm{d}y$$

Since \int y^{k}\mathrm{d}y=\frac{y^{k+1}}{k+1} for k\neq -1, replace \int y^{3}\mathrm{d}y with \frac{y^{4}}{4}. Multiply 8 times \frac{y^{4}}{4}.

$$\frac{y^{5}}{5}+2y^{4}+16\int y^{2}\mathrm{d}y$$

Since \int y^{k}\mathrm{d}y=\frac{y^{k+1}}{k+1} for k\neq -1, replace \int y^{2}\mathrm{d}y with \frac{y^{3}}{3}. Multiply 16 times \frac{y^{3}}{3}.

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

Simplify.

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

If F\left(y\right) is an antiderivative of f\left(y\right), then the set of all antiderivatives of f\left(y\right) is given by F\left(y\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

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

Differentiate w.r.t. y

$$\left(y\left(y+4\right)\right)^{2}$$