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

The integral of (1/(\sqrt[5]{x^2)}+1/x-x) is 5/3 x^{3/5}+ln|x|-(x^2)/2+C

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

\frac{5}{3} x^{\frac{3}{5}} + ln ∣ x ∣ - \frac{x ^{2}}{2} + C

Solution steps

assuming

x \geq 0

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= \frac{5}{3} x^{\frac{3}{5}} + ln ∣ x ∣ - \frac{x ^{2}}{2}

Add a constant to the solution

= \frac{5}{3} x^{\frac{3}{5}} + ln ∣ x ∣ - \frac{x ^{2}}{2} + C

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