What is the integral of 3/(sqrt(x^2-9)) ?

The integral of 3/(sqrt(x^2-9)) is 3ln(1/3 |sqrt(x^2-9)+x|)+C

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integral of 3/(sqrt(x^2-9))

\int \frac{3}{x ^{2} - 9} d x

3 ln (\frac{1}{3} x^{2} - 9 + x) + C

Solution steps

\int \frac{3}{x ^{2} - 9} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{x ^{2} - 9} d x

Apply Trigonometric Substitution

= 3 \cdot \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= 3 ln ∣ tan (u) + sec (u) ∣

Substitute back

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

= 3 ln tan (arcsec (\frac{1}{3} x)) + sec (arcsec (\frac{1}{3} x))

Simplify

3 ln tan (arcsec (\frac{1}{3} x)) + sec (arcsec (\frac{1}{3} x)) : 3 ln (\frac{1}{3} x^{2} - 9 + x)

= 3 ln (\frac{1}{3} x^{2} - 9 + x)

Add a constant to the solution

= 3 ln (\frac{1}{3} x^{2} - 9 + x) + C

What is the integral of 3/(sqrt(x^2-9)) ?
The integral of 3/(sqrt(x^2-9)) is 3ln(1/3 |sqrt(x^2-9)+x|)+C