What is the integral of 1/(x^5)-1/x ?

The integral of 1/(x^5)-1/x is -1/(4x^4)-ln|x|+C

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integral of 1/(x^5)-1/x

\int \frac{1}{x ^{5}} - \frac{1}{x} d x

- \frac{1}{4 x ^{4}} - ln ∣ x ∣ + C

Solution steps

\int \frac{1}{x ^{5}} - \frac{1}{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 ^{5}} d x - \int \frac{1}{x} d x

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

\int \frac{1}{x} d x = ln ∣ x ∣

= - \frac{1}{4 x ^{4}} - ln ∣ x ∣

Add a constant to the solution

= - \frac{1}{4 x ^{4}} - ln ∣ x ∣ + C

x^{^{''}} - 8 x^{^{'}} + 17 x = 0

\frac{d}{d x} (e^{\frac{x}{( x - 6 )}})

\frac{d}{d x} (ln (1 - \frac{3}{x}))

x \to 2 lim (\frac{x ^{2} - 2 x}{x})

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

What is the integral of 1/(x^5)-1/x ?
The integral of 1/(x^5)-1/x is -1/(4x^4)-ln|x|+C