Solve ∫ x(x^2+1) wrt x

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

Evaluate

$$\frac{x^{4}}{4}+\frac{x^{2}}{2}+С$$

Solution Steps

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

$$\int x^{3}+x\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{3}\mathrm{d}x+\int x\mathrm{d}x$$

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

$$\frac{x^{4}}{4}+\int x\mathrm{d}x$$

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

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

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

$$\frac{x^{4}}{4}+\frac{x^{2}}{2}+С$$

Differentiate w.r.t. x

$$x^{3}+x$$