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Question
Evaluate the following:
`int_0^1 x log(1 + 2x) "d"x`
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Solution
Let I = `int_0^1 x log(1 + 2x) "d"x`
= `[log (1 + 2x) x^2/2]_0^1 - int_0^1 2/(1 + 2x) x^2/2 "d"x` .....[Integrating by parts]
= `1/2 [x^2 log (1 + 2x)]_0^1 - int_0^1 x^2/(1 + 2x) "d"x`
= `1/2 [1 log 3 - 0] - int_0^1 (x/2 - x/(2(1 + 2x)))"d"x`
= ` 1/2 log 3 - 1/2 int_0^1 x "d"x + 1/2 int_0^1 x/(1 + 2x) "d"x`
= `1/2 log 3 - 1/2 [x^2/2]_0^1 + 1/4 int_0^1 ((2x + 1 - 1))/((2x + 1)) "d"x`
= `1/2 log 3 - 1/2 [1/2 - 0] + 1/4 int_0^1 "d"x - 1/4 int_0^1 1/(1 + 2x) "d"x`
= `1/2 log 3 - 1/4 + 1/4 - 1/8 [log (2x + 1)]_0^1`
= `1/2 log 3 - 1/4 + 1/4 - 1/8 [log 3 - log 1]`
= `1/2 log 3 - 1/8 log 3`
= `3/8 log 3`
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