Question

Evaluate the intergal

Answer 1

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

Answer 2

I suspect some creative manipulation will be required. I also see arc tan in this thing somehow.

Answer 3

you could just expand the bottom

Answer 4

Completing the square does not help.

Answer 5

How would that help?

Answer 6

ahh wait i was thinking the other way around lol

Answer 7

:) .

Answer 8

I do see myself using an arc tan in there though somehow.

Answer 9

I am not sure how or when though.

Answer 10

partials? you try that

Answer 11

Partials would make it even harder.

Answer 12

partial fractions you're thinking of correct?

Answer 13

Yes sir.

Answer 14

write denominator as an exponent to the negative power then apply product and chain rule.

Answer 15

@plate110 : that's for derivatives lol.

Answer 16

okay, integration by parts then.

Answer 17

@harsh314 That substitution does not help becuase I do not have it's derivative anywhere.

Answer 18

@plate110 :: Lol, I do not have products in this intergal. No IBP here.

Answer 19

ah i was going to say that

Answer 20

I don't like wolfram's method. It seems confusing and I am sure there is another way.

Answer 21

the method where you add constants to make your derivative

Answer 22

it's been a while since i've done that

Answer 23

I should metion the intergal I posted is the result of using partial fractions once already. THis is the "SImplified" Intergal lol.

Answer 24

well not add constants ... add and subtract to manipulate the equation

Answer 25

\[\int\limits\limits_{}^{}\frac{ 2x^4-4x^3+13x^2-6x+10 }{ (x-2)(x^2-2x+5)^2 }dx\]

Answer 26

i'd say do it wolfram's way i can't see it geting much simpler than that

Answer 27

That is the original intergal.

Answer 28

I used partial fractions to break it down into 3 simpler fractions.

Answer 29

I got it...I got it.....

Answer 30

My decomposition was correct as it matched wolfram's answer.

Answer 31

\[\int\limits\limits_{}^{}\frac{ 2 }{ x-2 }+\frac{ 4 }{ x^2-2x+5 }+\frac{ x }{ (x^2-2x+5)^2 }dx\]

Answer 32

This is the correct decomposition. I figured out the first two integrals but I cannot solve the last one.

Answer 33

alright so why don't you use wolfram's method

Answer 34

Seems confusing and hard >.> .

Answer 35

I really think the only two possible ways is to use their or partial fractions again

Answer 36

Hmm... Okay then :/ .

Answer 37

Thanks anyways :) .

Answer 38

it's notreally hard, when you see something that relates to your derivative you're looking for but is missing a constant is when you look at that specific way

Answer 39

you have x in the numerator but no 2 so you need to figure out how you can subtract 2 from the top only to add two back to it

Answer 40

Hmm... Wolfram can't decompose this...

Answer 41

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

Answer 42

but you also need to get a 2x in the top so you need to multiply both top and bottom by 2

Answer 43

so first you would have multiplied by 2/2 and then you'd add and subtract to get your derivative

Answer 44

Yeah I got that. I can't seem to use partial fractions again however so I guess I am going to have to go with WOlfram's method.

Answer 45

\[\int \frac{2(x-1+1)}{2(x^2-2x+5)^2}=\int \frac{2(x-1)}{2(x^2-2x+5)^2}+\frac{2}{2(x^2-2x+5)^2}\]

Answer 46

Thanks :) .