What is the integral of (x^4-\sqrt[3]{x})/(6sqrt(x)) ?

The integral of (x^4-\sqrt[3]{x})/(6sqrt(x)) is 1/135 (5x^{9/2}-27x^{5/6})+C

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integral of (x^4-\sqrt[3]{x})/(6sqrt(x))

\int \frac{x ^{4} - 3 x}{6 x} d x

\frac{1}{135} (5 x^{\frac{9}{2}} - 27 x^{\frac{5}{6}}) + C

Solution steps

\int \frac{x ^{4} - 3 x}{6 x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} \cdot \int \frac{x ^{4} - 3 x}{x} d x

Expand

\frac{x ^{4} - 3 x}{x} : x^{\frac{7}{2}} - \frac{1}{6 x}

= \frac{1}{6} \cdot \int x^{\frac{7}{2}} - \frac{1}{6 x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{6} (\int x^{\frac{7}{2}} d x - \int \frac{1}{6 x} d x)

\int x^{\frac{7}{2}} d x = \frac{2}{9} x^{\frac{9}{2}}

\int \frac{1}{6 x} d x = \frac{6}{5} x^{\frac{5}{6}}

= \frac{1}{6} (\frac{2}{9} x^{\frac{9}{2}} - \frac{6}{5} x^{\frac{5}{6}})

Simplify

\frac{1}{6} (\frac{2}{9} x^{\frac{9}{2}} - \frac{6}{5} x^{\frac{5}{6}}) : \frac{1}{135} (5 x^{\frac{9}{2}} - 27 x^{\frac{5}{6}})

= \frac{1}{135} (5 x^{\frac{9}{2}} - 27 x^{\frac{5}{6}})

Add a constant to the solution

= \frac{1}{135} (5 x^{\frac{9}{2}} - 27 x^{\frac{5}{6}}) + C

\int \frac{1}{x} ln (x) d x

\int - (t - 6)^{2} d t

\frac{d}{d y} (x^{0.5} y^{0.5})

x^{\frac{7}{9}}

f (x) = x^{2} - 7, at x = 2

What is the integral of (x^4-\sqrt[3]{x})/(6sqrt(x)) ?
The integral of (x^4-\sqrt[3]{x})/(6sqrt(x)) is 1/135 (5x^{9/2}-27x^{5/6})+C