Solve ∫ sqrt{t}+sqrt{t}dt

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

Evaluate

$$\frac{4t^{\frac{3}{2}}}{3}+С$$

Solution Steps

Integrate the sum term by term.

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

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

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

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

$$\frac{2t^{\frac{3}{2}}+2t^{\frac{3}{2}}}{3}$$

Simplify.

$$\frac{4t^{\frac{3}{2}}}{3}$$

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.

$$\frac{4t^{\frac{3}{2}}}{3}+С$$

Differentiate w.r.t. t

$$2\sqrt{t}$$