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

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

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

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

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

Solution steps

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

Apply Integral Substitution

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

Factor

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

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

Substitute back

u = \frac{x}{a}

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

Simplify

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

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

Add a constant to the solution

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

x \to 0 + lim (arctan (x))

\frac{d}{d x} (2 cos (x) - 3 sin (x) + 4)

\int arctan (6 t) d t

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

y^{^{''}} + 4 y^{^{'}} + 4 y = \frac{ln ( t )}{t} e^{- 2 t}

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