Solve ∫ (1+3t)t^2dt

$$\int ( 1 + 3 t ) t ^ { 2 } d t$$

Evaluate

$$\frac{3t^{4}}{4}+\frac{t^{3}}{3}+С$$

Solution Steps

Use the distributive property to multiply 1+3t by t^{2}.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

$$\frac{t^{3}}{3}+3\int t^{3}\mathrm{d}t$$

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

$$\frac{t^{3}}{3}+\frac{3t^{4}}{4}$$

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

$$\frac{t^{3}}{3}+\frac{3t^{4}}{4}+С$$

Differentiate w.r.t. t

$$\left(3t+1\right)t^{2}$$