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

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

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

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

\frac{3}{4} ln x^{4} - 1 + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 3 \cdot \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

= 3 \cdot \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{4} ln ∣ u ∣

Substitute back

u = x^{4} - 1

= 3 \cdot \frac{1}{4} ln x^{4} - 1

Simplify

3 \cdot \frac{1}{4} ln x^{4} - 1 : \frac{3}{4} ln x^{4} - 1

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

Add a constant to the solution

= \frac{3}{4} ln x^{4} - 1 + C

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