What is the integral of (x^4)/(x^5-8) ?

The integral of (x^4)/(x^5-8) is 1/5 ln|x^5-8|+C

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integral of (x^4)/(x^5-8)

\int \frac{x ^{4}}{x ^{5} - 8} d x

\frac{1}{5} ln x^{5} - 8 + C

Solution steps

\int \frac{x ^{4}}{x ^{5} - 8} d x

Apply u-substitution

= \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} - 8

= \frac{1}{5} ln x^{5} - 8

Add a constant to the solution

= \frac{1}{5} ln x^{5} - 8 + C

\int (cos (2 x))^{3} d x

\int \frac{4 + 6 x}{1 + x ^{2}} d x

\int \frac{1 + x}{x ^{2} + 1} d x

\int \frac{x ^{2} - 8 x + 1}{x ^{2} + 1} d x

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

What is the integral of (x^4)/(x^5-8) ?
The integral of (x^4)/(x^5-8) is 1/5 ln|x^5-8|+C