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Question
Evaluate the following definite integral:
`int_1^3 logx.dx`
Solution
Let I = `int_1^3 logx.dx`
= `int_1^3 logx.1.dx`
= `|logx int1.dx|_1^3 - int_1^3 [d/dx (logx) int1.dx].dx`
= `|logx.x int_1^3 - int_1^3 1/x. x. dx|`
= `|logx. x int_1^3 - int_1^3 1.dx|`
=`[x.log x]_1^3 - [x]_1^3`
= (3 . log 3 − 1 . log 1) − (3 − 1)
= (3 log 3 – 0) – 2
= 3 log 3 – 2
= log 33 – 2
∴ I = log 27 – 2
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