Solve ∫ (9/sqrt[4]{t}+4/t^7)dt

$$\int ( \frac { 9 } { \sqrt[ 4 ] { t } } + \frac { 4 } { t ^ { 7 } } ) d t$$

Evaluate

$$12t^{\frac{3}{4}}-\frac{2}{3t^{6}}+С$$

Solution Steps

Integrate the sum term by term.

$$\int \frac{9}{\sqrt[4]{t}}\mathrm{d}t+\int \frac{4}{t^{7}}\mathrm{d}t$$

Factor out the constant in each of the terms.

$$9\int \frac{1}{\sqrt[4]{t}}\mathrm{d}t+4\int \frac{1}{t^{7}}\mathrm{d}t$$

Rewrite \frac{1}{\sqrt[4]{t}} as t^{-\frac{1}{4}}. Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{-\frac{1}{4}}\mathrm{d}t with \frac{t^{\frac{3}{4}}}{\frac{3}{4}}. Simplify. Multiply 9 times \frac{4t^{\frac{3}{4}}}{3}.

$$12t^{\frac{3}{4}}+4\int \frac{1}{t^{7}}\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int \frac{1}{t^{7}}\mathrm{d}t with -\frac{1}{6t^{6}}. Multiply 4 times -\frac{1}{6t^{6}}.

$$12t^{\frac{3}{4}}-\frac{2}{3t^{6}}$$

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.

$$12t^{\frac{3}{4}}-\frac{2}{3t^{6}}+С$$

Differentiate w.r.t. t

$$\frac{9}{\sqrt[4]{t}}+\frac{4}{t^{7}}$$