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Ac3

  • one year ago

Evaluate the Integral of (x-3)/(x^2+2x+4)^2 please I'm stuck on this stupid problem.

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  1. alekos
    • one year ago
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    looks like a tough one, give me 5 minutes

  2. alekos
    • one year ago
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    Is this the integral? \[\int\limits \frac{ x-3 }{ (x ^{2} +2x+4)^2}\]

  3. Ac3
    • one year ago
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    yes

  4. Ac3
    • one year ago
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    answer in back of book is (-1/2(x^2+2x+4))-(2sqrt(3)/9)arctan((x+1)/sqrt3)-(2(x+1))/(3(x^2+2x+4))+C

  5. Ac3
    • one year ago
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    I'm going to sleep I'll be back in about 8-9 hours.

  6. IrishBoy123
    • one year ago
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    recognising that \(\large \frac{d}{dx} (x^2 + 2x + 4) = 2(x+1)\) you can start with : \(\large \int \frac{x-3}{(x^2 + 2x + 4)^2} \ dx= \int \frac{x+1}{(x^2 + 2x + 4)^2} - \frac{4}{(x^2 + 2x + 4)^2} \ dx\) so the first part falls out simply as \(\large \frac{-1/2}{(x^2 + 2x + 4)} \) for the second part, completing the square on the denominator leads to the rest: \( \large \int \frac{-4}{(x^2 + 2x + 4)^2} \ dx = \int \frac{-4}{((x + 1)^2 + \sqrt{3}^2)^2} \ dx\) which suggests a substitution \(\large x + 1 = \sqrt{3} tan \theta, \ dx = \sqrt{3} sec^2 \theta \ d \theta\) all of which simplifies eventually to: \( \frac{2 \sqrt{3}}{9} \int cos 2 \theta + 1 \ dx\) integrating plus some double angle formula plus this [from the \(\theta\) substitution]: |dw:1437734244042:dw| gets you home

  7. alekos
    • one year ago
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    good work @IrishBoy123. Heading in the right direction

  8. alekos
    • one year ago
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    by the way @Ac3 , this method leads to the answer that you've posted because I just completed the solution

  9. Ac3
    • one year ago
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    Thanks! Can't believe I didn't think of Trigonometric substitution.

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