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September 19, 2014

September 19, 2014

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

7e^(7t)

Divided By

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

Using partial fractions

- Calculus - Partial Fractions -
**Count Iblis**, Friday, March 14, 2008 at 9:27amSubstitute 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 -
**Sean**, Friday, March 14, 2008 at 9:41amThank you so much.

- Calculus - Partial Fractions -
**Sean**, Friday, March 14, 2008 at 9:49amI'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 -
**Count Iblis**, Friday, March 14, 2008 at 10:52amt = Log(x)/7 ---->

x = exp(7 t)

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