integration question?

Question

How do you integrate:

dy/dx= 1 / [x^2 - 1]

Answer 1

Partial fractions decomposition.

(1/(x^2 - 1)) = A/(x+1) + B/(x-1)
A= -1/2
B = 1/2

int(-1/(2(x+1))) + int(1/(2(x-1)))

-0.5ln|x+1| + 0.5ln|x-1| + C

Answer 2

if dy/dx= 1 / [x^2 - 1]
then y = 1 / [x^2 - 1]...
howver, the integral of y would then be
ln(x^2-1)/2x+C

Answer 3

You can substitute u = x^2-1 since you don't have the derivative of u. Simple fractions is the right way to do this

Ilusion

Answer 4

Make the substitution u = x²-1and then do it by repeated parts. You should get something like y = (ln(x-1) - ln(x+1))/2 + C
(I just kinda did that one 'quick and dirty' in my head, don't trust it too far ☺)

Doug