What is the integral of 1/(ssqrt(s^2-9)) ?

The integral of 1/(ssqrt(s^2-9)) is 1/3 arcsec(1/3 s)+C

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integral of 1/(ssqrt(s^2-9))

\int \frac{1}{s s ^{2} - 9} d s

\frac{1}{3} arcsec (\frac{1}{3} s) + C

Solution steps

\int \frac{1}{s s ^{2} - 9} d s

Apply Trigonometric Substitution

: \int \frac{1}{3} d u

= \int \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{1}{3} s)

= \frac{1}{3} \cdot 1 \cdot arcsec (\frac{1}{3} s)

Simplify

= \frac{1}{3} arcsec (\frac{1}{3} s)

Add a constant to the solution

= \frac{1}{3} arcsec (\frac{1}{3} s) + C

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What is the integral of 1/(ssqrt(s^2-9)) ?
The integral of 1/(ssqrt(s^2-9)) is 1/3 arcsec(1/3 s)+C