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posted by on .

How do you integrate:

1/(x^2-1) dx


  • Calculus - ,

    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)

    -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 - ,

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

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