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))

integral

\frac{x ^{3}}{( 1 - x ^{4} )}

- \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

sin^{2} (\frac{x}{2})

\frac{( 1 + x )}{( 1 + x ^{2} )}

y^{- 1}

f (x) = (x^{2} + 2 x + 5)^{\frac{1}{2}}

x \frac{cosh ( x )}{( sinh ( 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