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

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

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

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

\frac{1}{2 x} (ln \frac{y}{x} + 1 - ln \frac{y}{x} - 1) + C

Solution steps

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

Factor

x^{2} - y^{2} : - (- x^{2} + y^{2})

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

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Factor

u^{2} - 1 : - (- u^{2} + 1)

= - \frac{1}{x} \cdot \int \frac{1}{- ( - u ^{2} + 1 )} d u

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= - \frac{1}{x} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = \frac{y}{x}

= - \frac{1}{x} (- (\frac{ln \frac{y}{x} + 1}{2} - \frac{ln \frac{y}{x} - 1}{2}))

Simplify

- \frac{1}{x} (- (\frac{ln \frac{y}{x} + 1}{2} - \frac{ln \frac{y}{x} - 1}{2})) : \frac{1}{2 x} (ln \frac{y}{x} + 1 - ln \frac{y}{x} - 1)

= \frac{1}{2 x} (ln \frac{y}{x} + 1 - ln \frac{y}{x} - 1)

Add a constant to the solution

= \frac{1}{2 x} (ln \frac{y}{x} + 1 - ln \frac{y}{x} - 1) + C

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