What is the integral of (x^{1/3}-x^{1/4})/(4x^{1/2)} ?

The integral of (x^{1/3}-x^{1/4})/(4x^{1/2)} is 1/30 (9x^{5/6}-10x^{3/4})+C

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integral of (x^{1/3}-x^{1/4})/(4x^{1/2)}

\int \frac{x ^{\frac{1}{3}} - x ^{\frac{1}{4}}}{4 x ^{\frac{1}{2}}} d x

\frac{1}{30} (9 x^{\frac{5}{6}} - 10 x^{\frac{3}{4}}) + C

Solution steps

\int \frac{x ^{\frac{1}{3}} - x ^{\frac{1}{4}}}{4 x ^{\frac{1}{2}}} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{x ^{\frac{1}{3}} - x ^{\frac{1}{4}}}{x ^{\frac{1}{2}}} d x

Expand

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

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

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

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

= \frac{1}{4} (\frac{6}{5} x^{\frac{5}{6}} - \frac{4}{3} x^{\frac{3}{4}})

Simplify

\frac{1}{4} (\frac{6}{5} x^{\frac{5}{6}} - \frac{4}{3} x^{\frac{3}{4}}) : \frac{1}{30} (9 x^{\frac{5}{6}} - 10 x^{\frac{3}{4}})

= \frac{1}{30} (9 x^{\frac{5}{6}} - 10 x^{\frac{3}{4}})

Add a constant to the solution

= \frac{1}{30} (9 x^{\frac{5}{6}} - 10 x^{\frac{3}{4}}) + C

\int \frac{1}{2 + 2 cos ( x )} d x

\frac{d y}{d x} = x e^{x^{2} - y}

\int x^{3} \cdot e^{x^{2}} d x

\int \frac{8}{x} + 8 x d x

\frac{\partial}{\partial y} (\frac{x}{( x ^{3} + y ^{2} )})

What is the integral of (x^{1/3}-x^{1/4})/(4x^{1/2)} ?
The integral of (x^{1/3}-x^{1/4})/(4x^{1/2)} is 1/30 (9x^{5/6}-10x^{3/4})+C