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Find the roots, then you can factor it. Then you want to try partial fraction decomposition.
I am lost at completing the square.
\[\large x^2+bx\]To complete the square, we take half of the \(\large b\) term, and square it. \[\large x^2+bx+\left(\frac{b}{2}\right)^2\]Now we can't just add this number, it will change the value of our polynomial, we want to keep it balanced. So we have to also subtract this number. \[\large \color{royalblue}{x^2+bx+\left(\frac{b}{2}\right)^2}-\left(\frac{b}{2}\right)^2\]The reason for doing this is, the blue part will now factor down into a perfect square. A binomial containing x and half of the b term.\[\large \color{royalblue}{\left(x+\frac{b}{2}\right)^2}-\left(\frac{b}{2}\right)^2\]

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Let's see if we can apply this to our problem here.
\[\large x^2+x+1\]The \(\large b\) coefficient appears to be 1. (The middle term ~ because \(\large x\) is the same as \(\large 1\cdot x\) ). So to to complete the square, we'll take half of 1, and square it.\[\large \color{orangered}{x^2+x+\left(\frac{1}{2}\right)^2}-\left(\frac{1}{2}\right)^2+1\]I moved the 1 out of the way, we don't want to deal with that right now. Ok the orange part should give us a perfect square now,\[\large \color{orangered}{\left(x+\frac{1}{2}\right)^2}-\left(\frac{1}{2}\right)^2+1\] Which will simplify further to,\[\large \left(x+\frac{1}{2}\right)^2+\frac{3}{4}\]
\[4/3 \int\limits_{}^{}1/((4/3) u^2 +1) du\] What next?
(3/4)∫1/((4/3)u2+1)du I think this is right. What do I do next?
this seems much more easier than what is being done... but thats just me
\[\frac{4}{3}\int\frac{1}{\frac{4}{3}u^2+1}du=\frac{\frac{4}{3}}{\frac{4}{3}} \int \frac{1}{u^2+1}du\]
I agree. the book (answer in the back) and wolfram alpha I believe use a substitution here of 2/sqrt3 which gets complicated..
oh wait
is it \[\frac{1}{4/3(u^2+1)}\]
?
1/x2+x+1 tp begin with.
Does that answer your question?
what is problem... like that is in the book
without anything done to it
\[\int\limits_{}^{} (x^2 -x + 2) / (x^3 -1) \ \ dx\]
I understand the partial fraction decomposition element to this. I am still lost at the point mentioned to complete the square. I haven't been taught yet.
alright well whered the partials you used?
I ask this because if you use partials you should've gotten a natural log for one of the fractions
When you have \(u^2+1\) it's a sign to try the trig sub \(u=\tan\theta\).
yeah the answer so far is .. \[2/3 \ln (x+1) + 1/6 \ln (x^2 + x + 1) -9/2\int\limits_{}^{} dx / (x^2 + x + 1) \]
u = 4/sqrt3 ?
sorry 2/sqrt3 = u for the substitution?
What is the original integral?
∫(x2−x+2)/(x3−1) dx
the last part you have to
complete the square and use trig sub
\[x^2+x+1\] half of b = 1/2.. \[\frac{b}{4}^2=(\frac{b}{2})^2=\frac{1}{4}\]
if you add that you must subtract \[\frac{1}{x^2+x+\frac{1}{4}+1-\frac{1}{4}}\]
this is where he ot \[\frac{1}{(x+\frac{1}{2})^2+\frac{3}{4}}\]
then factor out the 3/4 in the bottom.
ok now you get to where you were before right?
right
:'(
now let \[x=\frac{a}{b}tan(\theta)\]
is a/b 3/4 or does should I do a substitution like 2/sqrt 3 ?

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