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Question
Solve the following : `int_0^1 (1)/(2x - 3)*dx`
Solution
Let I = `int_0^1 (1)/(2x - 3)*dx`
Put 2x – 3 =t
∴ 2·dx = dt
∴ dx = `"dt"/(2)`
When x = 0t = 2(0) – 3 = – 3
When x = 1, t = 2(1) – 3 = – 1
∴ I = `int_(-3)^(-1) (1)/"t"*"dt"/(2)`
= `(1)/(2) int_(-3)^(-1) "dt"/"t"`
= `(1)/(2)[log |"t"|]_(-3)^(-1)`
= `(1)/(2)[log|-1| - log|-3|]`
= `(1)/(2)(log 1 - log 3)`
= `(1)/(2)(0 - log 3)`
∴ I = `-(1)/(2) log 3`.
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