int. dx/ (x^2 +2x+10) using partial fraction. please.

- anonymous

int. dx/ (x^2 +2x+10) using partial fraction. please.

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- .Sam.

you can't factor that

- .Sam.

best approach is to complete the square and then trig substitution

- wasiqss

yehh, it will have complex factors

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## More answers

- anonymous

yup

- anonymous

@lilMissMindset To complete the square
x^2 +2x+10= x^2+2x+1+9 = (x+1)^2 +3^2
\[\int\limits_{}^{}\frac{dx}{(x+1)^2+3^2}\]
Now complete the rest

- anonymous

Remember
\[\int\limits_{}^{}\frac{dt}{x^2+a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a})\]

- anonymous

You can eaily prove the above by substituting x = atan(theta)

- anonymous

@shivam_bhalla she said partial fraction

- TuringTest

but it cannot be done with partial fraction if it cannot be factored

- wasiqss

traile , it willl become very complex then

- anonymous

yeah, im supposed to use that, partial fraction

- anonymous

@LOL, whjy do you want to complicate things when there is a aeasier method available. If you still insist on partial fraction, then it is fine

- wasiqss

lil miss it will only have complex factors

- TuringTest

there is either a typo, or it's impossible

- anonymous

*why

- TuringTest

I mean that using partial fractions is impossible... or at least redundant

- anonymous

what am i going to do about this problem then?

- anonymous

Take x+1 = t
dx=dt
You still get the same thing with partial fractions too

- TuringTest

how would you do this with partial fractions?
I don't see it... perhaps I am wrong though, it wouldn't be the first time :P

- anonymous

do you think quadratic factors can be use

- anonymous

\[\frac{1}{t^2+9} = \frac{At+B}{t^2+9}\]
we see a = 0, B=1
We again get back the same thing. So partail fraction approach should be useless

- anonymous

*partial

- TuringTest

like I said then, it's just redundant

- TuringTest

by impossible, I meant that the operation is useless, as you just said

- wasiqss

you fail turing xD lol jk

- TuringTest

@lilMissMindset are you \(sure\) there isn't a typo in your post?

- anonymous

yea. im quite sure i typed it right.

- anonymous

let's ignore the partial fraction thingy then.

- TuringTest

then shivam bhalla has shown you the right way
do you know trig substitution integrals?

- anonymous

yeah. i'm sorry, i'll do it using trig substitution. thank you so much.

- TuringTest

welcome!

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