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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.

Let's see if we can apply this to our problem here.

\[4/3 \int\limits_{}^{}1/((4/3) u^2 +1) du\]
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

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

\[\int\limits_{}^{} (x^2 -x + 2) / (x^3 -1) \ \ dx\]

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 \(u=\tan\theta\).

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 ?