Solve ∫ (t^2+1)dt

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

Evaluate

$$\frac{t^{3}}{3}+t+С$$

Solution Steps

Integrate the sum term by term.

$$\int t^{2}\mathrm{d}t+\int 1\mathrm{d}t$$

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}.

$$\frac{t^{3}}{3}+\int 1\mathrm{d}t$$

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}t=at.

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

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.

$$\frac{t^{3}}{3}+t+С$$

Differentiate w.r.t. t

$$t^{2}+1$$