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Posted by on Friday, March 14, 2008 at 8:54am.

What is the integral of

7e^(7t)

Divided By

e^14t+13e^7t+36

Using partial fractions

  • Calculus - Partial Fractions - , Friday, March 14, 2008 at 9:27am

    Substitute t = Log(x)/7

    Integral is then proportional to:

    Integral of dx/[x^2 + 13 x + 36] =

    Integral of dx/[(x+4)(x+9)]

    1/[(x+4)(x+9)] = A/(x+4) + B/(x+9)


    Multiply both sides by x+4 and take limit x to -4:

    1/5 = A

    Multiply both sides by x+9 and take limit x to -9:

    1/5 = A

    -1/5 = B

  • Calculus - Partial Fractions - , Friday, March 14, 2008 at 9:41am

    Thank you so much.

  • Calculus - Partial Fractions - , Friday, March 14, 2008 at 9:49am

    I've reached the last step of

    (7/5) ln((x+4)/(x+9)

    But how do I substitute the variable 't' back in?

  • Calculus - Partial Fractions - , Friday, March 14, 2008 at 10:52am

    t = Log(x)/7 ---->

    x = exp(7 t)

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