Question

how to do this integral?

Answer 1

\[\int\limits_{}\frac{ x^2 + 1 }{ (x-3)(x-2)^2}dx\]

Answer 2

HI!!

Answer 3

hello!!

Answer 4

how are you at partial fractions?

Answer 5

i understand it a bit (split into parts, right? A/B?) but don't know quite how to apply it!

Answer 6

it is going to suck
want to do it step by step, or cheat?

Answer 7

can you show me how to do it step by step?

Answer 8

\[\frac{A}{x-3}+\frac{B}{x-2}+\frac{C}{(x-2)^2}\]

Answer 9

okay! what next? :O

Answer 10

that means \[A(x-2)^2+B(x-3)(x-2)+C(x-3)=x^2+1\]

Answer 11

let me know if that is clear, then we can continue

Answer 12

yes:)

Answer 13

it is clear:) what happens next?

Answer 14

we can get C right away by putting \(x=2\)
because the A and B will drop out

Answer 15

if you put \(x=2\) you get \[C(2-3)=2^2+1\] or
\[-C=5\] making \(C=-5\)

Answer 16

got that?

Answer 17

yes:) sorry, OS connection was very poor!! what happens next?

Answer 18

get A by setting x = 3
same technique

Answer 19

okay, so then i get 3^2 + 1 = 10 ?

Answer 20

A=10?

Answer 21

\[A(3-2)^2+B(3-3)(3-2)+C(3-3)=3^2+1\]

\[\checkmark\]

Answer 22

ok!! what do i do next?

Answer 23

\[10(x-2)^2+B(x-3)(x-2)-5(x-3)=x^2+1\]

you need B so set x to anything really. but let's just go with x = 0.

Answer 24

okay! so B=1?

Answer 25

sure?

10(4) + B(-3)(-2) - 5(-3) = 1
6B = (1 -15 -40)????

Answer 26

ohh oops!! it equals -55!

Answer 27

-0.1111 = B?

Answer 28

@IrishBoy123 i have to leave for an appointment now, but will be back in a few hours!! is it okay if we continue then, or if i come back and see what you say?

Answer 29

\[\color{red}6B = -5 \color{red}4\]

Answer 30

sure

Answer 31

@IrishBoy123 I'm back!! can you please help me explain the rest?

Answer 32

so B = -9 right?

what happens next?

Answer 33

let's agree where we got to first

Answer 34

\[\frac{10}{x-3}-\frac{9}{x-2}-\frac{5}{(x-2)^2}\]

are we sure that's a good summary?

Answer 35

okay! so A=10, B=-9 and C=-5 ?

Answer 36

now you must do
\[\int \frac{10}{x-3}dx- \int \frac{9}{x-2}dx-\int\frac{5}{(x-2)^2} dx\]

3 separate integrals

Answer 37

ooh okay, how do i start off?

Answer 38

first one

\[\int \frac{10}{x-3}dx\]

Answer 39

okay, so we get 10lnx-3 ?

Answer 40

y
second?

Answer 41

so it is not that?

Answer 42

\[\int \frac{9}{x-2}dx\]

Answer 43

or it is?

Answer 44

first part is right, do the second now

Answer 45

ohhh i see now and so now second we get 9lnx-2 ?

Answer 46

and for the third, we get -5/(x-2) ?

Answer 47

yes. just make a note that we need to put the correct signs between the integrals righ at the end as it is ∫dx - ∫dx - ∫dx

last one

\[\int\frac{5}{(x-2)^2} dx\]

Answer 48

yes

put it all together now

Answer 49

don't bother writing it out if you are happy with it all

Answer 50

\[\int\limits_{}10lnx-3 dx - \int\limits_{}9lnx-2dx + \int\limits_{} \frac{ 5 }{ x-2 }dx\]

Answer 51

like that?

Answer 52

no, drop all the integration signs now that you have integrated

so
\[10 \ln(x-3) - 9\ln(x-2) + \frac{ 5 }{ x-2 }\]

Answer 53

OK?

Answer 54

ohh okay!! what now? :)

Answer 55

i think the fat lady is now singing.

Answer 56

hahaha ohh so we are done?

Answer 57

@IrishBoy123 ?

Answer 58

OMG!!!! we forgot something....

Answer 59

what?? :O

Answer 60

the integration constant!!

Answer 61

how embarassing after all that :-))

Answer 62

ohh + C ?

Answer 63

indeed!

Answer 64

ohh okie!! yay!! thank you so much!!

Answer 65

you and misty did all the heavy lifting, we just finished it off:)

good night!

Answer 66

good night!:D