What is the integral of (x^2+3x+1)/(\sqrt[5]{x^3)} ?

The integral of (x^2+3x+1)/(\sqrt[5]{x^3)} is 5/12 x^{12/5}+15/7 x^{7/5}+5/2 x^{2/5}+C

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integral of (x^2+3x+1)/(\sqrt[5]{x^3)}

\int \frac{x ^{2} + 3 x + 1}{5 x ^{3}} d x

\frac{5}{12} x^{\frac{12}{5}} + \frac{15}{7} x^{\frac{7}{5}} + \frac{5}{2} x^{\frac{2}{5}} + C

Solution steps

\int \frac{x ^{2} + 3 x + 1}{5 x ^{3}} d x

5 x^{3} = x^{\frac{3}{5}},

assuming

x \geq 0

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

Expand

\frac{x ^{2} + 3 x + 1}{x ^{\frac{3}{5}}} : x^{\frac{7}{5}} + 3 x^{\frac{2}{5}} + \frac{1}{x ^{\frac{3}{5}}}

= \int x^{\frac{7}{5}} + 3 x^{\frac{2}{5}} + \frac{1}{x ^{\frac{3}{5}}} 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 x^{\frac{7}{5}} d x + \int 3 x^{\frac{2}{5}} d x + \int \frac{1}{x ^{\frac{3}{5}}} d x

\int x^{\frac{7}{5}} d x = \frac{5}{12} x^{\frac{12}{5}}

\int 3 x^{\frac{2}{5}} d x = \frac{15}{7} x^{\frac{7}{5}}

\int \frac{1}{x ^{\frac{3}{5}}} d x = \frac{5}{2} x^{\frac{2}{5}}

= \frac{5}{12} x^{\frac{12}{5}} + \frac{15}{7} x^{\frac{7}{5}} + \frac{5}{2} x^{\frac{2}{5}}

Add a constant to the solution

= \frac{5}{12} x^{\frac{12}{5}} + \frac{15}{7} x^{\frac{7}{5}} + \frac{5}{2} x^{\frac{2}{5}} + C

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What is the integral of (x^2+3x+1)/(\sqrt[5]{x^3)} ?
The integral of (x^2+3x+1)/(\sqrt[5]{x^3)} is 5/12 x^{12/5}+15/7 x^{7/5}+5/2 x^{2/5}+C