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Well, for one thing you have written the natural log of the absolute value of 3-x.

Yeah I used long division initially to pull out this:
\[1 + \frac{-9}{x^3-3x^2}\]

Integral of one with respect to x is just x, logically

I think I mistyped my second post, sry @helder_edwin

ln|3-(4)| = ln|-1| = ln 1 = 0

That's a nice simplification

secondly, the logs should not contain the absolute value symbols around them

and the minuses cancel out inside the logs

*reading carefully*

Oh... properties of logs...
\[\ln|a|-\ln|b|=\ln|\frac{a}{b}|\]

yes - but you shouldn't use the '|' symbols here

as they usually indicate absolute value

I work it out to be exactly the same as the answer wolfram gives

I get:\[\frac{23}{20}+ln(\frac{5}{8})\approx 0.68\]

Ugh I see now I used the wrong base log on my calculator... /facepalm

Derp.

np :)

Thank you so much you were a great help! @TuringTest give that claymation-man a medal ^_^

you are more than welcome my friend! :)