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Question
Using second fundamental theorem, evaluate the following:
`int_1^"e" ("d"x)/(x(1 + logx)^3`
Sum
Solution
= `int_1^"e" (1 + logx)^-3/x "d"x`
= `[("f"(x)^(-3 + 1))/(-3 + 1)]_1^"e"`
= `[(1 + log x)^-2/-2]_1^"e"`
= `- 1/2 [[1 + log x]^-2]_1^"e"`
= `- 1/2 [(1 + log "e")^-2 (1 + log 1)^-2]`
= - 1/2 [(1 + 1)^-2 - (1)^-2]`
= `- 1/2 [1/(2)^2 - 1/(1)^2]`
= `- 1/2[1/4 - 1]`
= `-1/2[(1 - 4)/4]`
= `- 1/2[(-3)/4]`
= `3/8`
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Definite Integrals
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