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Question
Ok I should know this but I can't remember how too. I have
to do a long division on x^2/ x+4 and then do
integration. Can you help me with the first step of long
division. Thanks
I wouldn't bother with division. I would let u=x+4, then du = dx.
⌠ x²
⌡x+4 dx is the problem we have.
x = u-4, dx = du, x+4 = u, so now we have
⌠(u-4)²
⌡ u du
This reduces when the top is multiplied out. The top is really
u² - 8u + 16, and when this is divided by u, you get
u - 8 + 16/u.
When that is integrated, you get
u²/2 - 8u + 16*ln(u) + C.
Remembering that u=x+4, this is then
(x+4)² - 8(x+4) +16*ln(x+4) + C.
That can be reduced to
x²+8x+16 - 8x-32 + 16*ln(x+4).
It can be further reduced to
x² - 16 + 16*ln(x+4).
Of course, there should be a +C in there if you have no limits,
so it's really x² - 16 + 16*ln(x+4) + C.
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