calculate the integral e^x/(e^x-2)(e^x+5)

- megannicole51

calculate the integral e^x/(e^x-2)(e^x+5)

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- megannicole51

@pgpilot326

- megannicole51

@satellite73

- anonymous

you are going to give me a complex
first a u sub
then partial fractions

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## More answers

- anonymous

i.e. first compute the integral of
\[\frac{x}{(x-2)(x+5)}\] then replace \(x\) by \(e^x\)

- megannicole51

hahahah im sorry!!

- anonymous

i refuse to solve using linear systems
(just kidding)

- megannicole51

wait you can replace e^x with x??

- megannicole51

lol u know u love helping me!

- anonymous

ok if you don't like that, put \(u=e^x, du =e^xdx\) and get
\[\frac{u}{(u-2)(u+5)}\] if that is better on the eyes

- anonymous

i just replaced \(e^x\) by \(x\) instead of by \(u\) but no matter

- megannicole51

i like x better:)

- anonymous

i bet you can finish from here right? it is going to be another partial fraction problem, the only difference was the u - sub as a first step

- megannicole51

so when do i put back the e^x?

- anonymous

\[\frac{x}{(x-2)(x+5)}=\frac{A}{x-2}+\frac{B}{x+5}\]
at the very end, as any u - sub right?

- megannicole51

right!

- megannicole51

if i get it wrong then ill be back!!

- anonymous

u - sub is the first technique you learned i am sure

- anonymous

yeah but really really try to find A and B in your head
i am sure you can do it, even though they are fractions

- megannicole51

okay ill try!

- anonymous

try it now and let me know what you get

- anonymous

you two...

- megannicole51

okay i got A=(2/7) and B=(5/7) and its wrong....what did i do:(

- megannicole51

jk i know whats wrong!!! i didnt substitute it back!!

- megannicole51

nope its still wrong:( lol

- anonymous

no i think that is right

- anonymous

oh no i am so sorry, i steered you wrong

- anonymous

it is not
\[\frac{x}{(x-2)(x+5)}\] it is
\[\frac{1}{(x-2)(x+5)}\] damn damn damn

- megannicole51

(2/7)ln(abs(e^x-2))+(5/7)ln(abs(e^x+5))+c

- anonymous

many apologies, it must be getting late
i did the damn u sub wrong!!

- anonymous

\[u=e^x, du =e^xdx\] gives
\[\frac{1}{(x-2)(x+5)}\] not matter, we can fix it without any more work

- anonymous

just erase the 2 and the 5 in the numerators and replace them by 1

- anonymous

(1/7)ln(abs(e^x-2))+(1/7)ln(abs(e^x+5))+c

- anonymous

i am really sorry. all that work for nothing

- megannicole51

its still wrong:/

- megannicole51

no dont be sorry! im getting practice doing the steps!

- anonymous

ok hold the phone

- anonymous

\[\frac{1}{(x-2)(x+5)}=\frac{A}{x-2}+\frac{B}{x+5}\] put \(x=2\) get \(A+\frac{1}{7}\)

- anonymous

mean get \(A=\frac{1}{7}\)

- anonymous

put \(x=-5\) get \(B=-\frac{1}{7}\)

- anonymous

try
\[\frac{1}{7}\ln(e^x-2)-\frac{1}{7}\ln(e^x+5)\]

- anonymous

plus C and absolute values etc

- megannicole51

yay!!!!

- megannicole51

thank you thank you!!

- anonymous

whew
sorry
btw did you see how easy it is to find A and B? i mean the second time...

- megannicole51

yeah i did! i just need to practice more

- anonymous

good, do it, it is easy eventually

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