What is the integral of (x^4)/(x^5-2) ?

The integral of (x^4)/(x^5-2) is 1/5 ln|x^5-2|+C

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integral of (x^4)/(x^5-2)

\int \frac{x ^{4}}{x ^{5} - 2} d x

\frac{1}{5} ln x^{5} - 2 + C

Solution steps

\int \frac{x ^{4}}{x ^{5} - 2} d x

Apply u-substitution

: \int \frac{1}{5 u} d u

= \int \frac{1}{5 u} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{5} ln ∣ u ∣

Substitute back

u = x^{5} - 2

= \frac{1}{5} ln x^{5} - 2

Add a constant to the solution

= \frac{1}{5} ln x^{5} - 2 + C

\frac{d}{d x} (\frac{e ^{\frac{1}{x}}}{x ^{2}})

x \to \infty lim (1 - \frac{3}{x})

\frac{d}{d y} (- x y)

\frac{d y}{d x} = \frac{x ^{5}}{3 y ^{2}}

\frac{d}{d x} ((\frac{x}{9})^{2} + (\frac{y}{4})^{2}) = 1

What is the integral of (x^4)/(x^5-2) ?
The integral of (x^4)/(x^5-2) is 1/5 ln|x^5-2|+C