What is the integral of (27x^3-1)/(\sqrt[4]{x)} ?

The integral of (27x^3-1)/(\sqrt[4]{x)} is 36/5 x^{15/4}-4/3 x^{3/4}+C

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integral of (27x^3-1)/(\sqrt[4]{x)}

\int \frac{27 x ^{3} - 1}{4 x} d x

\frac{36}{5} x^{\frac{15}{4}} - \frac{4}{3} x^{\frac{3}{4}} + C

Solution steps

\int \frac{27 x ^{3} - 1}{4 x} d x

Expand

\frac{27 x ^{3} - 1}{4 x} : 27 x^{\frac{11}{4}} - \frac{1}{4 x}

= \int 27 x^{\frac{11}{4}} - \frac{1}{4 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

= \int 27 x^{\frac{11}{4}} d x - \int \frac{1}{4 x} d x

\int 27 x^{\frac{11}{4}} d x = \frac{36}{5} x^{\frac{15}{4}}

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

= \frac{36}{5} x^{\frac{15}{4}} - \frac{4}{3} x^{\frac{3}{4}}

Add a constant to the solution

= \frac{36}{5} x^{\frac{15}{4}} - \frac{4}{3} x^{\frac{3}{4}} + C

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What is the integral of (27x^3-1)/(\sqrt[4]{x)} ?
The integral of (27x^3-1)/(\sqrt[4]{x)} is 36/5 x^{15/4}-4/3 x^{3/4}+C