Solve = ∫ 2u(2-u^2)du

$$= \int 2 u ( 2 - u ^ { 2 } ) d u$$

Evaluate

$$-\frac{u^{4}}{2}+2u^{2}+С$$

Solution Steps

Use the distributive property to multiply 2u by 2-u^{2}.

$$\int 4u-2u^{3}\mathrm{d}u$$

Integrate the sum term by term.

$$\int 4u\mathrm{d}u+\int -2u^{3}\mathrm{d}u$$

Factor out the constant in each of the terms.

$$4\int u\mathrm{d}u-2\int u^{3}\mathrm{d}u$$

Since \int u^{k}\mathrm{d}u=\frac{u^{k+1}}{k+1} for k\neq -1, replace \int u\mathrm{d}u with \frac{u^{2}}{2}. Multiply 4 times \frac{u^{2}}{2}.

$$2u^{2}-2\int u^{3}\mathrm{d}u$$

Since \int u^{k}\mathrm{d}u=\frac{u^{k+1}}{k+1} for k\neq -1, replace \int u^{3}\mathrm{d}u with \frac{u^{4}}{4}. Multiply -2 times \frac{u^{4}}{4}.

$$2u^{2}-\frac{u^{4}}{2}$$

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

$$2u^{2}-\frac{u^{4}}{2}+С$$

Differentiate w.r.t. u

$$4u-2u^{3}$$