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

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

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

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

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

Solution steps

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (y ln (y^{2} - 2 x))

\int (x^{2} \cdot ln (x)) d x

\frac{1}{t ^{4}}

\int (sin (y) - y sin (x)) d x

\frac{\partial}{\partial t} (x + y sin (t))

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