Solve ∫ x^3+6x^2 wrt x

$$\int x ^ { 3 } + 6 x ^ { 2 } d x$$

Evaluate

$$\frac{x^{4}}{4}+2x^{3}+С$$

Solution Steps

Integrate the sum term by term.

$$\int x^{3}\mathrm{d}x+\int 6x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{3}\mathrm{d}x+6\int x^{2}\mathrm{d}x$$

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}.

$$\frac{x^{4}}{4}+6\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply 6 times \frac{x^{3}}{3}.

$$\frac{x^{4}}{4}+2x^{3}$$

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.

$$\frac{x^{4}}{4}+2x^{3}+С$$

Differentiate w.r.t. x

$$\left(x+6\right)x^{2}$$