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Evaluate the integral ; integral of (x^2+x+3)dx/ (x-1)^3
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\[\int\limits_{}^{} (x ^{2} + x + 3) dx / (x-1)^{3}\]
A/x-1 + B/(x-1)^2 + C/(x-1)^3... then
You're doing good, you want to know how to finish?
(x2+x+3)dx/(x−1)3 = A/x-1 + B/(x-1)^2 + C/(x-1)^3
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Multiply through by \[(x-1)^{3}\]
don't I have to get rid of the fractions first ?
Multiplying by \[(x-3)^{3}\] is a way of getting rid of the fractions.
jk, thats what it does haha YEAH
now find out the variables right?
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if x = 1, c = 5 but how do I find out the other variables
You substitute for 5 for C into the line \[x ^{2}+x+3=A(x-1)^{2}+B(x-1)+C\]
and expand
how do you expand ?
Multiply threw with B, expand \[(x-1)^{2}\]
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