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1. Find the integral of:

x^3/(x+1)(x+2)dx

Top heavy so I use the table method to divide through and find remainders:

I get x with remainder -3x^2 and -2x....

So I have A(x+2) + B(x+1) = -3x^2 -2x..

With a bit of magic I get:

-1/(x+1) + 8(x+2) and x

Integrates to:

0.5x^2 − ln |x + 1| + 8ln|x + 2| + c.

But apparently the answer is this and -3x...Where does the -3x come from?
2. Your "remainder" is wrong. You can divide (x+1)(x+2) into -3x^2 to get a smaller remainder.
3. For checking purposes may I recommend..

http://www72.wolframalpha.com/

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