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

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

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

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

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

Solution steps

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

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

Add a constant to the solution

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

\int \frac{6}{x ln ( 8 x )} d x

\int ln (x) x^{5} d x

\int (\frac{1}{x} - \frac{3}{16 x ^{2} + 1}) d x

\int (\frac{x}{x - 5}) d x

\int \frac{1}{x I n x} d x

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