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)

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

\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

x \to 2 lim ((x^{2})^{2})

y = x, y = 3 x, y = - x + 2

n = 0 \sum \infty n \cdot (\frac{3}{4})^{n}

\int \frac{( x + 1 )}{x ^{3} + x ^{2} + 6 x} d x

x^{^{'}} = 4 x - x^{3}

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