Evaluate ∫ 2x+1/x3+x2+2x+2dx


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Evaluate ∫ 2x+1/x3+x2+2x+2dx


Answer No.1


From: US
Posts: 16
Votes: 48
Start by partial fraction expansion. (2x+1)/(x^3 +3x^2+2x+2) = (1/3) (x/(x^2+2)) + (5/3) (1/(x^2+2))-(1/3)(1/(x+1)). Next integrate each of these separately. The third term has the integral (1/3) ln |x+1|. For the second one, we use the formula that the integral ofdx/(x^2+a^2) = (1/a) arctan(x/a). Here a= √2. For the first one, set u=x^2+2, then du=2xdx, i.e xdx=du/2. So we seek the integral of (1/3)(1/2) du/u which is (1/6) ln|u|=ln|x2+2|. Add these up to get the final answer. If your original integral means 2x+(1/x3)+x2+2x+2, then just integrate eachseparately using the fact that the integral of xn isxn+1/(n+1).

Answer No.2


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Integrate each portion separately, remembering the integrationrule: ∫ xn = (xn+1)/(n+1) +C. It may help to rewrite your integral as: ∫ 2x+x-3+x2+2x+2 dx. If you do everything correctly, you should get an answerof: 2x2-2x-2 + x3/3 + 2x + C

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