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Kelumptus

  • 3 years ago

Solution to the following integral (in post below)

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  1. Kelumptus
    • 3 years ago
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    \[\int\limits_{}^{} \frac{t-1}{t^{2}+4t+3} dt\] I'm kind of stuck on this one, i've tried a few different methods of u substitution but can't solve it.

  2. Callisto
    • 3 years ago
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    Does partial fraction help?

  3. Kelumptus
    • 3 years ago
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    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.

  4. Callisto
    • 3 years ago
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    Hmmm... \[\frac{t-1}{t^{2}+4t+3}=\frac{t-1}{(t+1)(t+3)} = \frac{A}{t+1}+\frac{B}{t+3}\] So, A(t+3) + B(t+1) = t-1 Comparing the coefficients, A+B = 1 3A +B = -1 Something like this?!

  5. Kelumptus
    • 3 years ago
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    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.

  6. Callisto
    • 3 years ago
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    Okay!!~

  7. Kelumptus
    • 3 years ago
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    Awesome, partial fractions was the key. Thanks for that =)

  8. Callisto
    • 3 years ago
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    You're welcome :)

  9. sehanhasan
    • 3 years ago
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    Use partial fraction method..

  10. shubhamsrg
    • 3 years ago
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    or you may use partial fractions alternatively, what i mean is t-1 / 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..

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