integral of 2sqrt(x)-\sqrt[3]{x}-x^4

Let's compute the integral:

\int (2 \sqrt{x} - \sqrt[3]{x} - x^{4}) d x

ⅰ

\int (2 \sqrt{x} - \sqrt[3]{x} - x^{4}) d x

\int 2 \sqrt{x} d x - \int \sqrt[3]{x} d x - \int x^{4} d x

\int 2 \sqrt{x} d x - \int \sqrt[3]{x} d x - \int x^{4} d x

We integrate

\sqrt[3]{x}

by applying:

\int x^{n} d x = \frac{x^{n + 1}}{n + 1}

ⓘ An antiderivative of x^n is x^(n+1)/(n+1), for n ≠ -1:

- \int \sqrt[3]{x} d x

- \frac{\sqrt[3]{x^{4}}}{\frac{4}{3}}

- \int x^{4} d x

- \frac{x^{5}}{5}

2 \cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}} - \frac{\sqrt[3]{x^{4}}}{\frac{4}{3}} - \frac{x^{5}}{5}

ⓘ Multiplying a fraction by a number multiplies its numerator:

2 \cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}}

\frac{2 x^{\frac{3}{2}}}{\frac{3}{2}}

\frac{2 x^{\frac{3}{2}}}{\frac{3}{2}} - \frac{\sqrt[3]{x^{4}}}{\frac{4}{3}} - \frac{x^{5}}{5}

ⓘ Dividing is multiplying by the reciprocal:

\frac{2 x^{\frac{3}{2}}}{\frac{3}{2}}

2 x^{\frac{3}{2}} \cdot \frac{2}{3}

- \frac{\sqrt[3]{x^{4}}}{\frac{4}{3}}

- \sqrt[3]{x^{4}} \cdot \frac{3}{4}

2 x^{\frac{3}{2}} \cdot \frac{2}{3} - \sqrt[3]{x^{4}} \cdot \frac{3}{4} - \frac{x^{5}}{5}

ⓘ Multiplications are done before additions and subtractions:

2 \cdot \frac{2}{3}

\frac{4}{3}

■Multiplying a fraction by a term multiplies its numerator:

\sqrt[3]{x^{4}} \cdot \frac{3}{4}

\frac{\sqrt[3]{x^{4}} \cdot 3}{4}

\frac{4}{3} \cdot x^{\frac{3}{2}} - \frac{\sqrt[3]{x^{4}} \cdot 3}{4} - \frac{x^{5}}{5}

ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{4}{3} \cdot x^{\frac{3}{2}}

\frac{4 x^{\frac{3}{2}}}{3}

\frac{4 x^{\frac{3}{2}}}{3} - \frac{\sqrt[3]{x^{4}} \cdot 3}{4} - \frac{x^{5}}{5}

▶Indefinite integrals are defined up to an additive constant C, so this yields:

- \frac{x^{5}}{5}

+ \frac{4 x^{\frac{3}{2}}}{3}

- \frac{3 \sqrt[3]{x^{4}}}{4}

+ C