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



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.Sam.
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you can't factor that

.Sam.
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best approach is to complete the square and then trig substitution

wasiqss
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yehh, it will have complex factors

traile
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yup

shivam_bhalla
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@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

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

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

traile
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@shivam_bhalla she said partial fraction

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

wasiqss
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traile , it willl become very complex then

lilMissMindset
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yeah, im supposed to use that, partial fraction

shivam_bhalla
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@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
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lil miss it will only have complex factors

TuringTest
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there is either a typo, or it's impossible

shivam_bhalla
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*why

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

lilMissMindset
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what am i going to do about this problem then?

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

TuringTest
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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

traile
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do you think quadratic factors can be use

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

shivam_bhalla
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*partial

TuringTest
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like I said then, it's just redundant

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

wasiqss
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you fail turing xD lol jk

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

lilMissMindset
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yea. im quite sure i typed it right.

lilMissMindset
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let's ignore the partial fraction thingy then.

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

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

TuringTest
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welcome!