Question

\(\huge \mathtt{\text{lalaly users only!}}\) \(\mathsf{\text{(for now :P)}}\)

integrate using partial fractions

\(\LARGE \int \frac{e^x}{(e^x - 1)(e^x + 3)} dx\)

Answer 1

@lalaly

Answer 2

lol sorry it took me a long time,,, i was solving another integral ...

Answer 3

nahh izfine that one was difficult :p haha

Answer 4

i made this post lalaly users only :p you should be flattered -__- haha jk

Answer 5

users ? are there more than 1 :)

Answer 6

exactly =))

Answer 7

@dumbcow why'd you reply here...now @lalaly is not going to answer my question :C

Answer 8

give me 5 more min the page crashed and lol i will do it dont worry,,, dumbcow is always welcome xD

Answer 9

oh..i thought you wont reply anymore coz you stopped typing :P hahaha

Answer 10

haha..ok i will delete

nevermind...thanks lalaly

Answer 11

let u=e^x
du=e^xdx

so the integral becomes\[\int\limits{\frac{du}{(u-1)(u+3)}}\]
first thing u do is break \[\frac{1}{(u-1)(u+3)}\]into partial fractions
\[\frac{1}{(u-1)(u+3)}=\frac{A}{u-1}+\frac{B}{u+3}\]
now you multiply both sides by (u-1)(u+3)
\[\cancel{(u-1)(u+3)}\times \frac{1}{\cancel{(u-1)(u+3)}}=(\frac{A}{u-1}+\frac{B}{u+3})\times(u-1)(u+3)\]
now we have
1=A(u+3)+B(u-1)
now to find A and B
from (u-1)(u+3)
u tkae first u=1
and substitute it in the equation
1=A(1+3)+B(1-1)
1=4A+0
so A=1/4

then u take u=-3
and u do the asme
1=A(-3+3)+B(-3-1)
1=0-4B
so B=-1/4

now that u know A and B
\[\frac{1}{(u-1)(u+3)}=\frac{\frac{1}{4}}{u-1}+\frac{\frac{-1}{4}}{u+3}\]
\[\frac{1}{(u-1)(u+3)}=\frac{1}{4(u-1)}-\frac{1}{4(u+3)}\]

Answer 12

now substitute that in the integral
\[\int\limits{\frac{du}{(u-1)(u+3)}}=\int\limits{(\frac{1}{4(u-1)}-\frac{1}{4(u+3)})du}\]

Answer 13

\[=\int\limits{\frac{du}{4(u-1)}}-\int\limits{\frac{du}{4(u+3)}}\]

Answer 14

now its an easy integral :D

Answer 15

tell me if there are things u dont understand:D

Answer 16

one more thing after u find the integral
u substitute u=e^x back xD

Answer 17

do it and i will tell u if its wrong or right

Answer 18

\(\large \frac{\ln (u-1)}{4} - \frac{\ln (u+3)}{4}\)

\(\Large \frac{\ln \frac{(u-1)}{(u+3)}}{4}\)

\(\LARGE \ln \sqrt[4]{\frac{(u-1)}{(u+3)}}\)

\(\LARGE \ln \sqrt[4]{\frac{(e^x - 1)}{(e^x + 3)}}\)

Answer 19

yeah +C xD

Answer 20

oh yeah that too :P

Answer 21

thanks :D that was pretty hard gotta practice more! then ill learn partial fractions + trig sub

Answer 22

cool im here if u need any more help

Answer 23

well...now that you mention it...

Answer 24

can you do this:

\(\LARGE \int \frac{x^4 - x^2 + 9}{x(x^2 + 3x + 3)^2} dx\)

looks hard :/ is it?

Answer 25

lol same way i showd
first thing factorise the denominator

Answer 26

can i try? :DD

Answer 27

\(\LARGE \frac{A}{x} + \frac{Bx + C}{x^2 -3x + 3} + \frac{Dx + E}{(x^2 - 3x + 3)^2} \)

that right?

Answer 28

the middle terms are +3x lol

Answer 29

yeah do the rest

Answer 30

A(x^2 + 3x + 3) + Bx + C(x^3 + 3x^2 + 3x) + Dx^2 + Ex = x^4 - x^2 + 9

Ax^2 + 3Ax + 3A + Bx^4 + 3Bx^3 + 3Bx^2 + Cx^3 + 3Cx^2 + 3Cx + Dx^2 + Ex = x^4 - x^2 + 9

Answer 31

3A = 9
A = 3

Answer 32

3A + 3C = 0
9 + 3C = 0
3C = -9
C = -3

Answer 33

coefficient of x^3

3B+ C = 0
3B - 3 = 0
3B = 3
B = 1

Answer 34

the previous ones were the constant and the coefficient of x respectively

Answer 35

coefficient of x^2

A + 3B + 3C + D = -1
3 + 3 - 9 + D = -1
D = -1 + 9 - 3 -3
D = 2

Answer 36

oh no! i forgot Exx im ruined T_T

Answer 37

lol solve it all and we'll see where u go wrong

Answer 38

okay :(