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

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

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

\int \frac{2 x}{( x ^{2} - 1 )} d x

ln x^{2} - 1 + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} - 1

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

Simplify

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

= ln x^{2} - 1

Add a constant to the solution

= ln x^{2} - 1 + C

f (x) = \frac{x ^{2}}{4 + x}

y^{^{'}} + 7 y = 9

\int (8 x ln (x)) d x

x \to \infty lim (3 x + 2)

\frac{d}{d y} (\frac{y}{2})

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