What is the integral of (x^3)/(81-x^4) ?

The integral of (x^3)/(81-x^4) is -1/4 ln|81-x^4|+C

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integral of (x^3)/(81-x^4)

\int \frac{x ^{3}}{81 - x ^{4}} d x

- \frac{1}{4} ln 81 - x^{4} + C

Solution steps

\int \frac{x ^{3}}{81 - x ^{4}} d x

Apply u-substitution

: \int - \frac{1}{4 u} d u

= \int - \frac{1}{4 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{4} ln ∣ u ∣

Substitute back

u = 81 - x^{4}

= - \frac{1}{4} ln 81 - x^{4}

Add a constant to the solution

= - \frac{1}{4} ln 81 - x^{4} + C

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What is the integral of (x^3)/(81-x^4) ?
The integral of (x^3)/(81-x^4) is -1/4 ln|81-x^4|+C