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Question
Evaluate `int_1^2 "e"^(2x) (1/x - 1/(2x^2)) "d"x`
Solution
Let I = `int_1^2 "e"^(2x) (1/x - 1/(2x^2)) "d"x`
= `int_1^2 "e"^(2x)*1/x "d"x - int_1^2 "e"^(2x)*1/(2x^2) "d"x`
= `[1/x int"e"^(2x) "d"x]_1^2 - int_1^2["d"/("d"x)(1/x)"f""e"^(2x) "d"x]"d"x - 1/2 int_1^2"e"^(2x)* 1/x^2 "d"x`
= `[1/x * ("e"^(2x))/2]_1^2 - int_1^2(-1/x^2)* ("e"^(2x))/2 "d"x - 1/2 int_1^2 "e"^(2x) * 1/x^2 "d"x`
= `(1/4 "e"^4 - "e"^2/2) + 1/2 int_1^2 "e"^(2x) * 1/x^2 "d"x - 1/2 int_1^2 "e"^(2x) * 1/x^2 "d"x`
∴ I = `"e"^4/4 - "e"^2/2`
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