How do I integrate 9/ (x^2-16) dx?

Answer 1

The solution is based on the chain rule:

d/dx ( f ( g(x) ) ) = f' ( g(x) ) * g'(x) )

First, with no constants of integration:

The integral of 1/x is ln(x), so the integral of 9 / (x² - 16) includes the term 9* ln (x² - 16)

Returning to the chain rule, we have g' = 2, so the original must include 1/2.

The postulated integral is then 9/2 * ( ln (x² - 16 ) + C

CHECK: (using the chain rule)

d/dx ( 9/2 * ( ln (x² - 16 ) + C ) = 9/2 * ( 1/(x² - 16) ) * 2 = 9 / (x² - 16) OK!!

Answer 2

Use the triangle method. Place sqrt(x²-16) on one leg, place 4 on the other leg, place x on the hypotenuse

Do the theta substitutions and you will end up with the integral of -(9/4)sec(theta)d(theta)

which gives -(9/4)ln(sec(theta) + tan(theta)) going back to the x domain you get

-(9/4)ln((x+4)/(sqrt(x²-16))