Solve ∫ (10sqrt[3]{t^2}+1/sqrt[3]{t^2})dt

$$\int ( 10 \sqrt[ 3 ] { t ^ { 2 } } + \frac { 1 } { \sqrt[ 3 ] { t ^ { 2 } } } ) d t$$

Evaluate

$$6t^{\frac{5}{3}}+3\sqrt[3]{t}+С$$

Solution Steps

Integrate the sum term by term.

$$\int 10t^{\frac{2}{3}}\mathrm{d}t+\int \frac{1}{t^{\frac{2}{3}}}\mathrm{d}t$$

Factor out the constant in each of the terms.

$$10\int t^{\frac{2}{3}}\mathrm{d}t+\int \frac{1}{t^{\frac{2}{3}}}\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{\frac{2}{3}}\mathrm{d}t with \frac{3t^{\frac{5}{3}}}{5}. Multiply 10 times \frac{3t^{\frac{5}{3}}}{5}.

$$6t^{\frac{5}{3}}+\int \frac{1}{t^{\frac{2}{3}}}\mathrm{d}t$$

Rewrite \frac{1}{t^{\frac{2}{3}}} as t^{-\frac{2}{3}}. Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{-\frac{2}{3}}\mathrm{d}t with \frac{t^{\frac{1}{3}}}{\frac{1}{3}}. Simplify and convert from exponential to radical form.

$$6t^{\frac{5}{3}}+3\sqrt[3]{t}$$

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.

$$6t^{\frac{5}{3}}+3\sqrt[3]{t}+С$$

Differentiate w.r.t. t

$$10t^{\frac{2}{3}}+\frac{1}{t^{\frac{2}{3}}}$$