What is the integral of

7e^(7t)

Divided By

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

Using partial fractions

4 answers

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
Thank you so much.
I've reached the last step of

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

But how do I substitute the variable 't' back in?
t = Log(x)/7 ---->

x = exp(7 t)