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

The integral of 1/(x^3*\sqrt[5]{1+1/x)} is -5/9 ((x+1)/x)^{9/5}+5/4 ((x+1)/x)^{4/5}+C

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

- \frac{5}{9} (\frac{x + 1}{x})^{\frac{9}{5}} + \frac{5}{4} (\frac{x + 1}{x})^{\frac{4}{5}} + C

Solution steps

Apply u-substitution

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

Expand

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

\int u^{\frac{4}{5}} d u = \frac{5}{9} u^{\frac{9}{5}}

= - (\frac{5}{9} u^{\frac{9}{5}} - \frac{5}{4} u^{\frac{4}{5}})

Substitute back

u = 1 + \frac{1}{x}

= - (\frac{5}{9} (1 + \frac{1}{x})^{\frac{9}{5}} - \frac{5}{4} (1 + \frac{1}{x})^{\frac{4}{5}})

Simplify

- (\frac{5}{9} (1 + \frac{1}{x})^{\frac{9}{5}} - \frac{5}{4} (1 + \frac{1}{x})^{\frac{4}{5}}) : - \frac{5}{9} (\frac{x + 1}{x})^{\frac{9}{5}} + \frac{5}{4} (\frac{x + 1}{x})^{\frac{4}{5}}

= - \frac{5}{9} (\frac{x + 1}{x})^{\frac{9}{5}} + \frac{5}{4} (\frac{x + 1}{x})^{\frac{4}{5}}

Add a constant to the solution

= - \frac{5}{9} (\frac{x + 1}{x})^{\frac{9}{5}} + \frac{5}{4} (\frac{x + 1}{x})^{\frac{4}{5}} + C

\frac{d}{d x} (ln (\frac{1}{2}))

f (x) = - 2 x^{\frac{2}{3}} - 4, at x = 8

\int \frac{x ^{3}}{1 - x ^{8}} d x

\frac{1 - 2 ln ( x )}{x ^{3}}

\int \frac{64 - x ^{2}}{x ^{2}} d x

What is the integral of 1/(x^3*\sqrt[5]{1+1/x)} ?
The integral of 1/(x^3*\sqrt[5]{1+1/x)} is -5/9 ((x+1)/x)^{9/5}+5/4 ((x+1)/x)^{4/5}+C