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Question
Evaluate `int_1^3 log x "d"x`
Solution
Let I = `int_1^3 log x "d"x`
= `int_1^3 logx*1 "d"x`
= `[log x int 1*"d"x]_1^3 - int_1^3["d"/("d"x) (log x) int1*"d"x] "d"x`
= `[logx*(x)]_1^3 - int_1^3 1/x*x "d"x`
= `[x log x]_1^3 - int_1^3 1*"d"x`
= (3 log 3 – 1 log 1) – `[x]_1^3`
= (3 log 3 – 0) – (3 – 1)
= 3 log 3 – 2
= log 33 – 2
∴ I = log 27 – 2
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