Solve ∫ (sqrt[4]{u^3}+sqrt[5]{4})du

$$\int ( \sqrt[ 4 ] { u ^ { 3 } } + \sqrt[ 5 ] { 4 } ) d u$$

Evaluate

$$\frac{4u^{\frac{7}{4}}}{7}+\sqrt[5]{4}u+С$$

Solution Steps

Integrate the sum term by term.

$$\int u^{\frac{3}{4}}\mathrm{d}u+\int \sqrt[5]{4}\mathrm{d}u$$

Since \int u^{k}\mathrm{d}u=\frac{u^{k+1}}{k+1} for k\neq -1, replace \int u^{\frac{3}{4}}\mathrm{d}u with \frac{4u^{\frac{7}{4}}}{7}.

$$\frac{4u^{\frac{7}{4}}}{7}+\int \sqrt[5]{4}\mathrm{d}u$$

Find the integral of \sqrt[5]{4} using the table of common integrals rule \int a\mathrm{d}u=au.

$$\frac{4u^{\frac{7}{4}}}{7}+\sqrt[5]{4}u$$

If F\left(u\right) is an antiderivative of f\left(u\right), then the set of all antiderivatives of f\left(u\right) is given by F\left(u\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$\frac{4u^{\frac{7}{4}}}{7}+\sqrt[5]{4}u+С$$

Differentiate w.r.t. u

$$u^{\frac{3}{4}}+\sqrt[5]{4}$$