<p>nahh… im not anti-civil rights… just anti-calculus…</p>
<p>can anyone help me integrate </p>
<p>y= 1/(x^2-2x+2)</p>
<p>so far i completed the square for the denominator</p>
<p>x^2-2x+2 = (x-1)^2 + 1
then i let v=(x-1)</p>
<p>int (1/(v^2+1))dv</p>
<p>what next… i think im on the right track and i know i should be ending up with arctan something or other… but i just cant get there for some reason… then again i havent slept for 31 hours… so thanks in advance!</p>
<p>ahh, i'm anti-calculus too!!! is there some kind of movement we can join to show our discontent?!?! lol</p>
<p>i'd help you..but...i kinda failed calc (more or less) >_<; </p>
<p><em>ah, i love how cc is like a homework help site lol</em></p>
<p>lol wait im soo stupid.. i really need to get some sleep..</p>
<p>so </p>
<p>int [1/(1+x^2)] = arctan(x) + C</p>
<p>so letting u=x-1</p>
<p>int [1/(u^2+1)) = arctan(u) + C</p>
<p>= arctan(x-1) + C</p>
<p>correct?</p>
<p>I don't know how to integrate something like this (yet our math class just finished 'u' substitution and the chapter on integration--are we supposed to learn all of that arctan stuff later or something?!). I put the function into my calculator so it can integrate it. I can't tell you how to do it, but the answer turns out to be:</p>
<p>[pi(arctan(x-1))]/180</p>
<p>In case you want to check your answer or something. Hope this helps!</p>
<p>.017453 arctan (x-1)</p>
<p>arctan(x-1) + C sounds right.</p>
<p>That's an inverse trig function for integration.
Says that int dx/(x^2+a^2) = (1/a) * arctan (x/a) + C</p>
<p>Can't you just use partial fraction decomposition here?</p>
<p>jpps1: What you mean partial fractions? Not really worth trying (the denominator isn't looking too promising for something like that. What would you break it up into? Not the best method here). And the numerator only has one item so you can't split it into multiple fractions with the same denominator. I would do:</p>
<p>y= 1/(x^2-2x+2)
y= 1/(x^2-2x + 1 + 1)
y= 1/((x-1)^2 + 1)
set u = x-1 so du = dx
integral of 1/((x-1)^2 + 1) dx
integral of 1/(u^2 + 1) du
This just equates to arctan(u) + C, so arctan(x - 1) + C.</p>
<p>Some things you just have to know (inverse trig identities, hyperbolic functions, standard trig functions, logarithmic, exponential, etc). I can show you why arctan is indeed the integral of the 1/blahblah if you want, but it's not something you should have to "figure out" on a test -- it's just something you should probably memorize. Kinda sad that a lot of stuff we do in math is just memory recall and not too much actual "problem-solving" (then again, harder integrals do take some effort).</p>
<p>I didn't really take a look at the problem, but yeah, you're right--that denominator doesn't work for partial fractions.</p>