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

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

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integral of a/(s^2-a^2)

\int \frac{a}{s ^{2} - a ^{2}} d s

- \frac{1}{2} ln \frac{s}{a} + 1 + \frac{1}{2} ln \frac{s}{a} - 1 + C

Solution steps

\int \frac{a}{s ^{2} - a ^{2}} d s

Take the constant out:

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

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

Apply Integral Substitution

= a \cdot \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

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

Factor

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

= a \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

= a \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}

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

Substitute back

u = \frac{s}{a}

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

Simplify

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

= - \frac{1}{2} ln \frac{s}{a} + 1 + \frac{1}{2} ln \frac{s}{a} - 1

Add a constant to the solution

= - \frac{1}{2} ln \frac{s}{a} + 1 + \frac{1}{2} ln \frac{s}{a} - 1 + C

\frac{d}{d x} (I n x)

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

\int \frac{1}{u ^{2} + u} d u

\frac{( s + 2 )}{s ( s + 4 )}

\frac{d}{d x} (tan (5 x + 5))

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