joesherman 0 Posted April 28, 2016 Share Posted April 28, 2016 Find integral from 0 to 2 of: X^{2} ^{ _______________} X^{2 }- 2X + 2 Link to post Share on other sites

ajb 1567 Posted April 28, 2016 Share Posted April 28, 2016 (edited) Is this homework? If so, what have you done so far to try to solve this? I can give you a hint if you are really stuck. Edited April 28, 2016 by ajb Link to post Share on other sites

joesherman 0 Posted April 28, 2016 Author Share Posted April 28, 2016 No, it isn't some assignment - I'm a teacher trying to stay on top of it; but -as everyone knows - I'm also victim of the same conspiracy of the churches oppressing & smearing the rock & rollers for decades, & this makes it so that I don't always have any professor or fellow colleague I can bounce thing off of. I've tried integration by parts, substitution (symbolic), partial fractions integration; I also tried completing the square - & there may be something going in that direction. I already solved another similar problem, finding the integral of X ________________ X^{2} - 2X + 2 Any instruction from the "Expert" would be awesome! Link to post Share on other sites

mathematic 104 Posted April 29, 2016 Share Posted April 29, 2016 Let Y=X-1. You now have an integrand (Y+1)^{2}/(Y^{2}+1)=1+2Y/(Y^{2}+1). (Note dY=dX) The indefinite integral is Y + ln(Y^{2}+1). Final answer 2(1+ln(2)). Link to post Share on other sites

ajb 1567 Posted April 29, 2016 Share Posted April 29, 2016 (edited) The method I would use in this example us to think about the derivative of the log of your deniminator. You see that this is almost what you are looking for. But you can then make a simple 'correction' to get the indefinite integral you require (trying adding 1 as this correction) You can then substitute in the value 0 and 2 to get your definite integral. P.S. You can always check the answer using Wolfram Alpha. Edited April 29, 2016 by ajb Link to post Share on other sites

Juno 13 Posted May 19, 2016 Share Posted May 19, 2016 (edited) I used the following:- [math] \int \frac{x^2}{x^2-2x+2} dx = \int \frac{x^2-2x+2+2x-2}{x^2-2x+2} dx = \int 1+\frac{2x-2}{x^2-2x+2} dx [/math] Then the integration of the fraction is straightforward as the top is the derivative of the bottom. Edited May 19, 2016 by Juno Link to post Share on other sites

deesuwalka 1 Posted November 16, 2016 Share Posted November 16, 2016 (edited) [latex] \int\limits^2_0 \dfrac{x^2}{x^2-2x+2} dx [/latex] [latex] =\int\limits^2_0 \dfrac{x^2-2x+2+2x-2}{x^2-2x+2} dx [/latex] [latex] =\int\limits^2_0 1+ \dfrac{2x-2}{x^2-2x+2} dx [/latex] [latex] =\int\limits^2_0 dx+\int\limits^2_0 \dfrac{2x-2}{x^2-2x+2} dx [/latex] Let [latex] t=x^2-2x+2 [/latex] [latex] dt=(2x-2)dx [/latex] [latex] =\bigg[x\bigg]^2_0+\int\limits^2_0\dfrac{dt}{t} [/latex] [latex] =2+\bigg[\ln\,t\bigg]^2_0\;\;\implies\bigg[\ln(x^2-2x+2)\bigg]^2_0 [/latex] [latex] =2+\bigg[\ln\,2-\ln\,2\bigg] [/latex] [latex] =2+0=2 [/latex] Edited November 16, 2016 by deesuwalka Link to post Share on other sites

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