What is the integral of (x^4)/(3-x^5) ?

The integral of (x^4)/(3-x^5) is -1/5 ln|3-x^5|+C

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integral of (x^4)/(3-x^5)

\int \frac{x ^{4}}{3 - x ^{5}} d x

- \frac{1}{5} ln 3 - x^{5} + C

Solution steps

\int \frac{x ^{4}}{3 - 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 = 3 - x^{5}

= - \frac{1}{5} ln 3 - x^{5}

Add a constant to the solution

= - \frac{1}{5} ln 3 - x^{5} + C

\frac{d}{d x} (\frac{a x + b}{a x - b})

f (x) = x^{2} + 4, (3, 13)

\int e^{- x} d x

\frac{\partial}{\partial x} (ln (2))

32 - 2 x^{2}, [0, 7]

What is the integral of (x^4)/(3-x^5) ?
The integral of (x^4)/(3-x^5) is -1/5 ln|3-x^5|+C