Solve ∫ 2u(u^2+1)du=

$$\int 2 u ( u ^ { 2 } + 1 ) d u =$$

Evaluate

$$\frac{u^{4}}{2}+u^{2}+С$$

Solution Steps

Use the distributive property to multiply 2u by u^{2}+1.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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 2 times \frac{u^{2}}{2}.

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

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.

$$\frac{u^{4}}{2}+u^{2}+С$$

Differentiate w.r.t. u

$$2u\left(u^{2}+1\right)$$