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

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

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

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

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

Solution steps

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

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

Add a constant to the solution

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

t^{2} y^{^{''}} + 2 t y^{^{'}} - 30 y = 0

x \to 3 lim (\frac{( x + 3 )}{x - 3})

\int (ln (5 x))^{2} d x

\int \frac{( 17 x + 3 )}{( 4 x + 1 ) ( x - 1 )} d x

x \to 0 lim (x^{2} ln (x))

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