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KelumptusBest ResponseYou've already chosen the best response.0
\[\int\limits_{}^{} \frac{t1}{t^{2}+4t+3} dt\] I'm kind of stuck on this one, i've tried a few different methods of u substitution but can't solve it.
 one year ago

CallistoBest ResponseYou've already chosen the best response.4
Does partial fraction help?
 one year ago

KelumptusBest ResponseYou've already chosen the best response.0
Yeah i think that might be the answer. I have only done partial fractions in the form where there is a coefficient and a variable however. Not when there are two terms in the numerator.
 one year ago

CallistoBest ResponseYou've already chosen the best response.4
Hmmm... \[\frac{t1}{t^{2}+4t+3}=\frac{t1}{(t+1)(t+3)} = \frac{A}{t+1}+\frac{B}{t+3}\] So, A(t+3) + B(t+1) = t1 Comparing the coefficients, A+B = 1 3A +B = 1 Something like this?!
 one year ago

KelumptusBest ResponseYou've already chosen the best response.0
Yep... i think i jumped the gun when i gave up on trying to solve with partial fractions. I just revised my notes and i'll have a go using partial fractions now. I'll let you know how i go.
 one year ago

KelumptusBest ResponseYou've already chosen the best response.0
Awesome, partial fractions was the key. Thanks for that =)
 one year ago

sehanhasanBest ResponseYou've already chosen the best response.0
Use partial fraction method..
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.1
or you may use partial fractions alternatively, what i mean is t1 / t^2 +4t +3 = 1/2 ( 2t  2)/(t^2 +4t +3) =1/2 ( 2t +4 6)/(t^2 +4t +3) = 1/2 ( 2t +4)/(t^2 +4t +3)  1/2 (6/(t^2 +4t +3)) integrate separately, for 1st part, if you let denominator =z,. you directly get numerator = dz .. we are concerned with integrating 1/(t^2 +4t +3) that's also simple, we see we have 1/(t+1)(t+3) = 1/2 (2 / (t+1)(t+3)) =1/2 ( t+3  (t+1))/(t+1)(t+3)) =>now when you separate out, you'll directly be able to integrate ..hope that helps.. note that i've used partial fractions only,but in a different manner..
 one year ago
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