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

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

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

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

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

Solution steps

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

Apply Integral Substitution

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

Factor

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

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

Substitute back

u = \frac{x}{y}

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

Simplify

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

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

Add a constant to the solution

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

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