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\[\int\limits x-1+\frac{ 1 }{ x+2 } +\frac{ 1 }{ x-1 } dx\]

is this the same as:

\[\int\limits x-1 dx +\int\limits \frac{ 1 }{ x+2 } dx +\int\limits \frac{ 1 }{ x-1 } dx\]
????

Yes, that would be correct.

You may also split (x - 1) into separate integrals, that should work out as well.

are these answers the same:

\[x^2 -2x +1 +2\ln \left| x+2 \right| +2\ln \left| x-1 \right|\]

and

\[\frac{ 1 }{ 2 }(x-2)x +\ln (x-1) +\ln (x+2)\]

??

The first one is my answer the second is wolfram alpha

and multiply through by two to clear the fraction?

I meant times two

= x²/2 - x + ln | x+2| + ln | x -1| + C

or x² - 2x + 2ln | x+2| + 2ln | x -1| + C

You have 4 separates terms to take derivative. The result looks neat without fraction !

*separated

I am still somewhat confused I don't get how they are the same

The first one has fraction 1/2
If you don't want fraction form, you need to multiply all terms by 2!

= x²/2 - x + ln | x+2| + ln | x -1| + C
x² - 2x + 2ln | x+2| + 2ln | x -1| + C

I got (x^2)-2x+1 and I just cannot see how that is the same as (x-2)x

This is just plain straight forward formula from the textbook!

" I got (x^2)-2x+1 "
Where do you get 1?

sorry I must have missed the "plain straightforward" part of my calculus II textbook

lol

Does it mean you understand your mistake?

u = x -1 --> du = dx
=> ∫ du/ u = ln u = ln | x-1|

it was just x-1 not 1/(x-1)

Oh, so they're 2 individual terms!

* to integrate