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

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

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integral of u/(u^2-1)

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

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

Solution steps

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

Apply u-substitution

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

= \int \frac{1}{2 v} d v

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}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} ln ∣ v ∣

Substitute back

v = u^{2} - 1

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (sin (4 x - 6 y))

f (x) = 6 sin (2 x) - 3 cos (3 x)

y^{^{''}} - y = t e^{t}

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