Express the function π(π₯) = π₯ /(π₯β3)2 as partial fractions and hence find β« π(π₯)πx
Solution
The function π(π₯) = π₯ /(π₯β3)2Β may be represented as sum of partial fractions f(x) = A/(x-3)+B/(x-3)2Β where A, B are arbitrary constants. From equality π₯ /(π₯β3)2 = A/(x-3)+B/(x-3)2 weβll get Β
A(x-3)+B = xΒ => A=1, B=3A=3 => π(π₯) = 1/(π₯β3)+3/(π₯β3)2 Β Β Β
Now β« π(π₯)πx = β« [1/(π₯β3)+3/(π₯β3)2]πx = β« [1/(π₯β3)]πx + 3β« [1/(π₯β3)2]πx = ln|x-3|-3/(π₯β3)+CΒ Β
Answer
π(π₯) = 1/(π₯β3)+3/(π₯β3)2
β« π(π₯)πx = ln|x-3|-3/(π₯β3)+C
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