Question

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

Answer 1

@pgpilot326

Answer 2

@satellite73

Answer 3

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

Answer 4

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

Answer 5

hahahah im sorry!!

Answer 6

i refuse to solve using linear systems
(just kidding)

Answer 7

wait you can replace e^x with x??

Answer 8

lol u know u love helping me!

Answer 9

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

Answer 10

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

Answer 11

i like x better:)

Answer 12

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

Answer 13

so when do i put back the e^x?

Answer 14

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

Answer 15

right!

Answer 16

if i get it wrong then ill be back!!

Answer 17

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

Answer 18

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

Answer 19

okay ill try!

Answer 20

try it now and let me know what you get

Answer 21

you two...

Answer 22

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

Answer 23

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

Answer 24

nope its still wrong:( lol

Answer 25

no i think that is right

Answer 26

oh no i am so sorry, i steered you wrong

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

i am really sorry. all that work for nothing

Answer 34

its still wrong:/

Answer 35

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

Answer 36

ok hold the phone

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

plus C and absolute values etc

Answer 42

yay!!!!

Answer 43

thank you thank you!!

Answer 44

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

Answer 45

yeah i did! i just need to practice more

Answer 46

good, do it, it is easy eventually