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

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

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

integral

\frac{1}{( a ^{2} - x ^{2} )}

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

Solution steps

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

Factor

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

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

Take the constant out:

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

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

\frac{x ^{2}}{( 1 - x )}

\frac{1}{( x ^{2} + a ^{2} ) ^{2}}

lo g_{10} (1 + x)

(x + y)^{2}

\frac{k \cdot x}{( 1 + x ^{2} )}

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