Solve ∫ (4/sqrt[3]{t}+3/t^6)dt

$$\int ( \frac { 4 } { \sqrt[ 3 ] { t } } + \frac { 3 } { t ^ { 6 } } ) d t$$

Evaluate

$$6t^{\frac{2}{3}}-\frac{3}{5t^{5}}+С$$

Solution Steps

Integrate the sum term by term.

$$\int \frac{4}{\sqrt[3]{t}}\mathrm{d}t+\int \frac{3}{t^{6}}\mathrm{d}t$$

Factor out the constant in each of the terms.

$$4\int \frac{1}{\sqrt[3]{t}}\mathrm{d}t+3\int \frac{1}{t^{6}}\mathrm{d}t$$

Rewrite \frac{1}{\sqrt[3]{t}} as t^{-\frac{1}{3}}. Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{-\frac{1}{3}}\mathrm{d}t with \frac{t^{\frac{2}{3}}}{\frac{2}{3}}. Simplify. Multiply 4 times \frac{3t^{\frac{2}{3}}}{2}.

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

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int \frac{1}{t^{6}}\mathrm{d}t with -\frac{1}{5t^{5}}. Multiply 3 times -\frac{1}{5t^{5}}.

$$6t^{\frac{2}{3}}-\frac{\frac{3}{t^{5}}}{5}$$

Simplify.

$$6t^{\frac{2}{3}}-\frac{3}{5t^{5}}$$

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.

$$6t^{\frac{2}{3}}-\frac{3}{5t^{5}}+С$$

Differentiate w.r.t. t

$$\frac{4}{\sqrt[3]{t}}+\frac{3}{t^{6}}$$