Asked by Sean
What is the integral of
7e^(7t)
Divided By
e^14t+13e^7t+36
Using partial fractions
7e^(7t)
Divided By
e^14t+13e^7t+36
Using partial fractions
Answers
Answered by
Count Iblis
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
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
Answered by
Sean
Thank you so much.
Answered by
Sean
I've reached the last step of
(7/5) ln((x+4)/(x+9)
But how do I substitute the variable 't' back in?
(7/5) ln((x+4)/(x+9)
But how do I substitute the variable 't' back in?
Answered by
Count Iblis
t = Log(x)/7 ---->
x = exp(7 t)
x = exp(7 t)
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.