What is the integral of ((10)/(y^{1/4)}+sqrt(y)) ?

The integral of ((10)/(y^{1/4)}+sqrt(y)) is 40/3 y^{3/4}+2/3 y^{3/2}+C

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integral of ((10)/(y^{1/4)}+sqrt(y))

\int (\frac{10}{y ^{\frac{1}{4}}} + y) d y

\frac{40}{3} y^{\frac{3}{4}} + \frac{2}{3} y^{\frac{3}{2}} + C

Solution steps

\int \frac{10}{y ^{\frac{1}{4}}} + y 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

= \int \frac{10}{y ^{\frac{1}{4}}} d y + \int y d y

\int \frac{10}{y ^{\frac{1}{4}}} d y = \frac{40}{3} y^{\frac{3}{4}}

\int y d y = \frac{2}{3} y^{\frac{3}{2}}

= \frac{40}{3} y^{\frac{3}{4}} + \frac{2}{3} y^{\frac{3}{2}}

Add a constant to the solution

= \frac{40}{3} y^{\frac{3}{4}} + \frac{2}{3} y^{\frac{3}{2}} + C

What is the integral of ((10)/(y^{1/4)}+sqrt(y)) ?
The integral of ((10)/(y^{1/4)}+sqrt(y)) is 40/3 y^{3/4}+2/3 y^{3/2}+C