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

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

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integral of y/(y^2-1)

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

\frac{1}{2} ln y^{2} - 1 + C

Solution steps

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

Apply u-substitution

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

= \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = y^{2} - 1

= \frac{1}{2} ln y^{2} - 1

Add a constant to the solution

= \frac{1}{2} ln y^{2} - 1 + C

sin (cos (tan (x)))

\frac{\partial}{\partial y} (\frac{x}{y + 2 z})

n = 1 \sum \infty (- 0.75)^{n}

\int x (x - 5) d x

y = 2 x^{2} ln (x), y = 8 ln (x)

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