What is the integral of (2x^2+4x^4-1)/(\sqrt[3]{x)} ?

The integral of (2x^2+4x^4-1)/(\sqrt[3]{x)} is 3/4 x^{8/3}+6/7 x^{14/3}-3/2 x^{2/3}+C

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integral of (2x^2+4x^4-1)/(\sqrt[3]{x)}

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

\frac{3}{4} x^{\frac{8}{3}} + \frac{6}{7} x^{\frac{14}{3}} - \frac{3}{2} x^{\frac{2}{3}} + C

Solution steps

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

Expand

\frac{2 x ^{2} + 4 x ^{4} - 1}{3 x} : 2 x^{\frac{5}{3}} + 4 x^{\frac{11}{3}} - \frac{1}{3 x}

= \int 2 x^{\frac{5}{3}} + 4 x^{\frac{11}{3}} - \frac{1}{3 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 2 x^{\frac{5}{3}} d x + \int 4 x^{\frac{11}{3}} d x - \int \frac{1}{3 x} d x

\int 2 x^{\frac{5}{3}} d x = \frac{3}{4} x^{\frac{8}{3}}

\int 4 x^{\frac{11}{3}} d x = \frac{6}{7} x^{\frac{14}{3}}

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

= \frac{3}{4} x^{\frac{8}{3}} + \frac{6}{7} x^{\frac{14}{3}} - \frac{3}{2} x^{\frac{2}{3}}

Add a constant to the solution

= \frac{3}{4} x^{\frac{8}{3}} + \frac{6}{7} x^{\frac{14}{3}} - \frac{3}{2} x^{\frac{2}{3}} + C

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What is the integral of (2x^2+4x^4-1)/(\sqrt[3]{x)} ?
The integral of (2x^2+4x^4-1)/(\sqrt[3]{x)} is 3/4 x^{8/3}+6/7 x^{14/3}-3/2 x^{2/3}+C