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

The integral of (y^{1/2}+y^{1/3})/(2y^{1/4)} is 2/65 (15y^{13/12}+13y^{5/4})+C

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integral of (y^{1/2}+y^{1/3})/(2y^{1/4)}

\int \frac{y ^{\frac{1}{2}} + y ^{\frac{1}{3}}}{2 y ^{\frac{1}{4}}} d y

\frac{2}{65} (15 y^{\frac{13}{12}} + 13 y^{\frac{5}{4}}) + C

Solution steps

\int \frac{y ^{\frac{1}{2}} + y ^{\frac{1}{3}}}{2 y ^{\frac{1}{4}}} d y

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{y ^{\frac{1}{2}} + y ^{\frac{1}{3}}}{y ^{\frac{1}{4}}} d y

Expand

\frac{y ^{\frac{1}{2}} + y ^{\frac{1}{3}}}{y ^{\frac{1}{4}}} : y^{\frac{1}{4}} + y^{\frac{1}{12}}

= \frac{1}{2} \cdot \int y^{\frac{1}{4}} + y^{\frac{1}{12}} d y

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}{2} (\int y^{\frac{1}{4}} d y + \int y^{\frac{1}{12}} d y)

\int y^{\frac{1}{4}} d y = \frac{4}{5} y^{\frac{5}{4}}

\int y^{\frac{1}{12}} d y = \frac{12}{13} y^{\frac{13}{12}}

= \frac{1}{2} (\frac{4}{5} y^{\frac{5}{4}} + \frac{12}{13} y^{\frac{13}{12}})

Simplify

\frac{1}{2} (\frac{4}{5} y^{\frac{5}{4}} + \frac{12}{13} y^{\frac{13}{12}}) : \frac{2}{65} (15 y^{\frac{13}{12}} + 13 y^{\frac{5}{4}})

= \frac{2}{65} (15 y^{\frac{13}{12}} + 13 y^{\frac{5}{4}})

Add a constant to the solution

= \frac{2}{65} (15 y^{\frac{13}{12}} + 13 y^{\frac{5}{4}}) + C

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What is the integral of (y^{1/2}+y^{1/3})/(2y^{1/4)} ?
The integral of (y^{1/2}+y^{1/3})/(2y^{1/4)} is 2/65 (15y^{13/12}+13y^{5/4})+C