Solve ∫ sqrt{6}tdt

$$\int \sqrt { 6 } t d t$$

Evaluate

$$\frac{\sqrt{6}t^{2}}{2}+С$$

Solution Steps

Factor out the constant using \int af\left(t\right)\mathrm{d}t=a\int f\left(t\right)\mathrm{d}t.

$$\sqrt{6}\int t\mathrm{d}t$$

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

$$\sqrt{6}\times \frac{t^{2}}{2}$$

Simplify.

$$\frac{\sqrt{6}t^{2}}{2}$$

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.

$$\frac{\sqrt{6}t^{2}}{2}+С$$

Differentiate w.r.t. t

$$\sqrt{6}t$$