What is the integral of (6x)/(x^2-1) ?

The integral of (6x)/(x^2-1) is 3ln|x^2-1|+C

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integral of (6x)/(x^2-1)

\int \frac{6 x}{x ^{2} - 1} d x

3 ln x^{2} - 1 + C

Solution steps

\int \frac{6 x}{x ^{2} - 1} d x

Take the constant out:

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

= 6 \cdot \int \frac{x}{x ^{2} - 1} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

= 6 \cdot \int \frac{1}{2 u} d u

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} - 1

= 6 \cdot \frac{1}{2} ln x^{2} - 1

Simplify

6 \cdot \frac{1}{2} ln x^{2} - 1 : 3 ln x^{2} - 1

= 3 ln x^{2} - 1

Add a constant to the solution

= 3 ln x^{2} - 1 + C

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What is the integral of (6x)/(x^2-1) ?
The integral of (6x)/(x^2-1) is 3ln|x^2-1|+C