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Mathematics 21 Online
OpenStudy (anonymous):

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

sam (.sam.):

you can't factor that

sam (.sam.):

best approach is to complete the square and then trig substitution

OpenStudy (wasiqss):

yehh, it will have complex factors

OpenStudy (anonymous):

yup

OpenStudy (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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

@shivam_bhalla she said partial fraction

OpenStudy (turingtest):

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

OpenStudy (wasiqss):

traile , it willl become very complex then

OpenStudy (anonymous):

yeah, im supposed to use that, partial fraction

OpenStudy (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

OpenStudy (wasiqss):

lil miss it will only have complex factors

OpenStudy (turingtest):

there is either a typo, or it's impossible

OpenStudy (anonymous):

*why

OpenStudy (turingtest):

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

OpenStudy (anonymous):

what am i going to do about this problem then?

OpenStudy (anonymous):

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

OpenStudy (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

OpenStudy (anonymous):

do you think quadratic factors can be use

OpenStudy (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

OpenStudy (anonymous):

*partial

OpenStudy (turingtest):

like I said then, it's just redundant

OpenStudy (turingtest):

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

OpenStudy (wasiqss):

you fail turing xD lol jk

OpenStudy (turingtest):

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

OpenStudy (anonymous):

yea. im quite sure i typed it right.

OpenStudy (anonymous):

let's ignore the partial fraction thingy then.

OpenStudy (turingtest):

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

OpenStudy (anonymous):

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

OpenStudy (turingtest):

welcome!

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