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Ac3
 one year ago
Evaluate the Integral of (x3)/(x^2+2x+4)^2 please I'm stuck on this stupid problem.
Ac3
 one year ago
Evaluate the Integral of (x3)/(x^2+2x+4)^2 please I'm stuck on this stupid problem.

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alekos
 one year ago
Best ResponseYou've already chosen the best response.0looks like a tough one, give me 5 minutes

alekos
 one year ago
Best ResponseYou've already chosen the best response.0Is this the integral? \[\int\limits \frac{ x3 }{ (x ^{2} +2x+4)^2}\]

Ac3
 one year ago
Best ResponseYou've already chosen the best response.0answer 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

Ac3
 one year ago
Best ResponseYou've already chosen the best response.0I'm going to sleep I'll be back in about 89 hours.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3recognising that \(\large \frac{d}{dx} (x^2 + 2x + 4) = 2(x+1)\) you can start with : \(\large \int \frac{x3}{(x^2 + 2x + 4)^2} \ dx= \int \frac{x+1}{(x^2 + 2x + 4)^2}  \frac{4}{(x^2 + 2x + 4)^2} \ dx\) 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

alekos
 one year ago
Best ResponseYou've already chosen the best response.0good work @IrishBoy123. Heading in the right direction

alekos
 one year ago
Best ResponseYou've already chosen the best response.0by the way @Ac3 , this method leads to the answer that you've posted because I just completed the solution

Ac3
 one year ago
Best ResponseYou've already chosen the best response.0Thanks! Can't believe I didn't think of Trigonometric substitution.
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