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

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

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

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

\frac{1}{5} ln x^{5} - 4 + C

Solution steps

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

= \frac{1}{5} ln x^{5} - 4

Add a constant to the solution

= \frac{1}{5} ln x^{5} - 4 + C

\frac{d}{d x} (\frac{4}{3} π \cdot x^{3})

y = ln (5 x^{2} + 7 y^{2})

\int_{- 1}^{9} (9 - x^{2} + 8 x) d x

\int \frac{x}{49 + x ^{4}} d x

\frac{2}{t} - \frac{4}{t ^{2}}

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