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Question
Evaluate `int_1^3 x^2*log x "d"x`
Solution
Let I = `int_1^3 x^2*log x "d"x`
= `[log x int x^2 "d"x]_1^3 - int_1^3["d"/("d"x)(log x) intx^2 "d"x]"d"x`
= `[log x* x^3/3]_1^3 - int_1^3 1/x*x^3/3 "d"x`
= `[9log3 - log1*1/3] - 1/3 int_1^3 x^2 "d"x`
= `(9log 3 - 0) - 1/3 [x^3/3]_1^3`
= `9log3 - 1/3(27/3 - 1/3)`
= `9log3 - 1/3(26/3)`
∴ I = `9log 3 - 26/9`
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