how to do this integral?

- anonymous

how to do this integral?

- schrodinger

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- anonymous

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

- misty1212

HI!!

- anonymous

hello!!

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## More answers

- misty1212

how are you at partial fractions?

- anonymous

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

- misty1212

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

- anonymous

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

- misty1212

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

- anonymous

okay! what next? :O

- misty1212

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

- misty1212

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

- anonymous

yes:)

- anonymous

it is clear:) what happens next?

- misty1212

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

- misty1212

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

- misty1212

got that?

- anonymous

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

- IrishBoy123

get A by setting x = 3
same technique

- anonymous

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

- anonymous

A=10?

- IrishBoy123

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

- anonymous

ok!! what do i do next?

- IrishBoy123

\[10(x-2)^2+B(x-3)(x-2)-5(x-3)=x^2+1\]
you need B so set x to anything really. but let's just go with x = 0.

- anonymous

okay! so B=1?

- IrishBoy123

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

- anonymous

ohh oops!! it equals -55!

- anonymous

-0.1111 = B?

- anonymous

@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?

- IrishBoy123

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

- IrishBoy123

sure

- anonymous

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

- anonymous

so B = -9 right?
what happens next?

- IrishBoy123

let's agree where we got to first

- IrishBoy123

\[\frac{10}{x-3}-\frac{9}{x-2}-\frac{5}{(x-2)^2}\]
are we sure that's a good summary?

- anonymous

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

- IrishBoy123

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

- anonymous

ooh okay, how do i start off?

- IrishBoy123

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

- anonymous

okay, so we get 10lnx-3 ?

- IrishBoy123

y
second?

- anonymous

so it is not that?

- IrishBoy123

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

- anonymous

or it is?

- IrishBoy123

first part is right, do the second now

- anonymous

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

- anonymous

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

- IrishBoy123

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

- IrishBoy123

yes
put it all together now

- IrishBoy123

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

- anonymous

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

- anonymous

like that?

- IrishBoy123

no, drop all the integration signs now that you have integrated
so
\[10 \ln(x-3) - 9\ln(x-2) + \frac{ 5 }{ x-2 }\]

- IrishBoy123

OK?

- anonymous

ohh okay!! what now? :)

- IrishBoy123

i think the fat lady is now singing.

- anonymous

hahaha ohh so we are done?

- anonymous

- IrishBoy123

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

- anonymous

what?? :O

- IrishBoy123

the integration constant!!

- IrishBoy123

how embarassing after all that :-))

- anonymous

ohh + C ?

- IrishBoy123

indeed!

- anonymous

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

- IrishBoy123

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

- anonymous

good night!:D

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