What is the integral of (3x^4-5x^3+7x-1)/(x^{3/4)} ?

The integral of (3x^4-5x^3+7x-1)/(x^{3/4)} is 12/17 x^{17/4}-20/13 x^{13/4}+28/5 x^{5/4}-4x^{1/4}+C

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integral of (3x^4-5x^3+7x-1)/(x^{3/4)}

\int \frac{3 x ^{4} - 5 x ^{3} + 7 x - 1}{x ^{\frac{3}{4}}} d x

\frac{12}{17} x^{\frac{17}{4}} - \frac{20}{13} x^{\frac{13}{4}} + \frac{28}{5} x^{\frac{5}{4}} - 4 x^{\frac{1}{4}} + C

Solution steps

\int \frac{3 x ^{4} - 5 x ^{3} + 7 x - 1}{x ^{\frac{3}{4}}} d x

Expand

\frac{3 x ^{4} - 5 x ^{3} + 7 x - 1}{x ^{\frac{3}{4}}} : 3 x^{\frac{13}{4}} - 5 x^{\frac{9}{4}} + 7 x^{\frac{1}{4}} - \frac{1}{x ^{\frac{3}{4}}}

= \int 3 x^{\frac{13}{4}} - 5 x^{\frac{9}{4}} + 7 x^{\frac{1}{4}} - \frac{1}{x ^{\frac{3}{4}}} 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 3 x^{\frac{13}{4}} d x - \int 5 x^{\frac{9}{4}} d x + \int 7 x^{\frac{1}{4}} d x - \int \frac{1}{x ^{\frac{3}{4}}} d x

\int 3 x^{\frac{13}{4}} d x = \frac{12}{17} x^{\frac{17}{4}}

\int 5 x^{\frac{9}{4}} d x = \frac{20}{13} x^{\frac{13}{4}}

\int 7 x^{\frac{1}{4}} d x = \frac{28}{5} x^{\frac{5}{4}}

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

= \frac{12}{17} x^{\frac{17}{4}} - \frac{20}{13} x^{\frac{13}{4}} + \frac{28}{5} x^{\frac{5}{4}} - 4 x^{\frac{1}{4}}

Add a constant to the solution

= \frac{12}{17} x^{\frac{17}{4}} - \frac{20}{13} x^{\frac{13}{4}} + \frac{28}{5} x^{\frac{5}{4}} - 4 x^{\frac{1}{4}} + C

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What is the integral of (3x^4-5x^3+7x-1)/(x^{3/4)} ?
The integral of (3x^4-5x^3+7x-1)/(x^{3/4)} is 12/17 x^{17/4}-20/13 x^{13/4}+28/5 x^{5/4}-4x^{1/4}+C