What is the integral of (y^2+4y-4)/(\sqrt[4]{y)} ?

The integral of (y^2+4y-4)/(\sqrt[4]{y)} is 4/11 y^{11/4}+16/7 y^{7/4}-16/3 y^{3/4}+C

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integral of (y^2+4y-4)/(\sqrt[4]{y)}

\frac{4}{11} y^{\frac{11}{4}} + \frac{16}{7} y^{\frac{7}{4}} - \frac{16}{3} y^{\frac{3}{4}} + C

Solution steps

Expand

Apply the Sum Rule:

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

\int y^{\frac{7}{4}} d y = \frac{4}{11} y^{\frac{11}{4}}

\int 4 y^{\frac{3}{4}} d y = \frac{16}{7} y^{\frac{7}{4}}

= \frac{4}{11} y^{\frac{11}{4}} + \frac{16}{7} y^{\frac{7}{4}} - \frac{16}{3} y^{\frac{3}{4}}

Add a constant to the solution

= \frac{4}{11} y^{\frac{11}{4}} + \frac{16}{7} y^{\frac{7}{4}} - \frac{16}{3} y^{\frac{3}{4}} + C

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What is the integral of (y^2+4y-4)/(\sqrt[4]{y)} ?
The integral of (y^2+4y-4)/(\sqrt[4]{y)} is 4/11 y^{11/4}+16/7 y^{7/4}-16/3 y^{3/4}+C