How do you integrate:

Question

1/(x^2-1) dx

?

Answer 1

draw a right triangle, x as the hypotensue, 1 as the adjacent side to angle theta, and the opposite side is sqrt (x^2-1)

costheta=1/x
-sintheta* dT=-1/x^2 dx or

dx= x^2 sinTheta dT

the integral is now...

INt 1/tanTheta*x^2 sinTheta dT
INT cosT/SinT* sec^2T*sinT dT
INT= secT dT=ln(secT+tanT)
= ln(x+x^2-1)

check all that, I did it in my head.

Answer 2

Hm, I've never done integrals using that method before. I'll check it.