megannicole51
  • megannicole51
calculate the integral e^x/(e^x-2)(e^x+5)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
megannicole51
  • megannicole51
@pgpilot326
megannicole51
  • megannicole51
@satellite73
anonymous
  • anonymous
you are going to give me a complex first a u sub then partial fractions

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anonymous
  • anonymous
i.e. first compute the integral of \[\frac{x}{(x-2)(x+5)}\] then replace \(x\) by \(e^x\)
megannicole51
  • megannicole51
hahahah im sorry!!
anonymous
  • anonymous
i refuse to solve using linear systems (just kidding)
megannicole51
  • megannicole51
wait you can replace e^x with x??
megannicole51
  • megannicole51
lol u know u love helping me!
anonymous
  • 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
  • anonymous
i just replaced \(e^x\) by \(x\) instead of by \(u\) but no matter
megannicole51
  • megannicole51
i like x better:)
anonymous
  • 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
  • megannicole51
so when do i put back the e^x?
anonymous
  • 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
  • megannicole51
right!
megannicole51
  • megannicole51
if i get it wrong then ill be back!!
anonymous
  • anonymous
u - sub is the first technique you learned i am sure
anonymous
  • 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
  • megannicole51
okay ill try!
anonymous
  • anonymous
try it now and let me know what you get
anonymous
  • anonymous
you two...
megannicole51
  • megannicole51
okay i got A=(2/7) and B=(5/7) and its wrong....what did i do:(
megannicole51
  • megannicole51
jk i know whats wrong!!! i didnt substitute it back!!
megannicole51
  • megannicole51
nope its still wrong:( lol
anonymous
  • anonymous
no i think that is right
anonymous
  • anonymous
oh no i am so sorry, i steered you wrong
anonymous
  • anonymous
it is not \[\frac{x}{(x-2)(x+5)}\] it is \[\frac{1}{(x-2)(x+5)}\] damn damn damn
megannicole51
  • megannicole51
(2/7)ln(abs(e^x-2))+(5/7)ln(abs(e^x+5))+c
anonymous
  • anonymous
many apologies, it must be getting late i did the damn u sub wrong!!
anonymous
  • 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
  • anonymous
just erase the 2 and the 5 in the numerators and replace them by 1
anonymous
  • anonymous
(1/7)ln(abs(e^x-2))+(1/7)ln(abs(e^x+5))+c
anonymous
  • anonymous
i am really sorry. all that work for nothing
megannicole51
  • megannicole51
its still wrong:/
megannicole51
  • megannicole51
no dont be sorry! im getting practice doing the steps!
anonymous
  • anonymous
ok hold the phone
anonymous
  • anonymous
\[\frac{1}{(x-2)(x+5)}=\frac{A}{x-2}+\frac{B}{x+5}\] put \(x=2\) get \(A+\frac{1}{7}\)
anonymous
  • anonymous
mean get \(A=\frac{1}{7}\)
anonymous
  • anonymous
put \(x=-5\) get \(B=-\frac{1}{7}\)
anonymous
  • anonymous
try \[\frac{1}{7}\ln(e^x-2)-\frac{1}{7}\ln(e^x+5)\]
anonymous
  • anonymous
plus C and absolute values etc
megannicole51
  • megannicole51
yay!!!!
megannicole51
  • megannicole51
thank you thank you!!
anonymous
  • anonymous
whew sorry btw did you see how easy it is to find A and B? i mean the second time...
megannicole51
  • megannicole51
yeah i did! i just need to practice more
anonymous
  • anonymous
good, do it, it is easy eventually

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