What is the integral of (x^4)/(1-x^5) ?

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

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integral of (x^4)/(1-x^5)

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

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

Solution steps

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

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

Add a constant to the solution

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

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

\int \frac{1}{400 - x ^{2}} d x

t (x) v^{''} + v^{'} = 0

\int \frac{3}{x ln ( 5 x )} d x

y^{10} \frac{d y}{d x} = - 9 cos (π x)

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