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August 1, 2014

August 1, 2014

Posted by **Kyle** on Wednesday, January 9, 2013 at 7:48pm.

1/(x^2-1) dx

?

- Calculus -
**bobpursley**, Wednesday, January 9, 2013 at 7:56pmdraw 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.

- Calculus -
**Kyle**, Wednesday, January 9, 2013 at 8:17pmHm, I've never done integrals using that method before. I'll check it.

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