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Question
`int (1 + x)/(x + "e"^(-x)) "d"x`
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Solution
Let I = `int (1 + x)/(x + "e"^(-x)) "d"x`
= `int (1 + x)/(x + 1/"e"^x) "d"x`
= `int (1 + x)/((x*"e"^x + 1)/("e"^x)) "d"x`
= `int ("e"^x (1 + x))/(x*"e"^x + 1) "d"x`
Put x. ex + 1 = t
∴ [x. (ex) + ex. (1) + 0] dx = dt
∴ ex (x + 1) dx = dt
∴ I = `int "dt"/"t"`
= log |t| + c
∴ I = log |x. ex + 1| + c
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