What is the integral of (x^3)/(1-x^4) ?

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

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integral of (x^3)/(1-x^4)

\int \frac{x ^{3}}{1 - x ^{4}} d x

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

Solution steps

\int \frac{x ^{3}}{1 - 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 = 1 - x^{4}

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

Add a constant to the solution

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

x = - 5, x = 1, y = 11 x, y = x^{2} - 12

2 x^{4}

\frac{s + 4}{s ^{2} + 6 s + 25}

(ln (3) + e^{x^{e}})^{e}

x \to - 1 + lim (1 - x^{2})

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