Question

`int_1^3 ("d"x)/(x(1 + logx)^2)` = ______.

Options

• log 2

• `1/(1 + log2)`

• `log2/(1 + log2)`

• `- 1/(1 + log2)`

Answer 1

`int_1^3 ("d"x)/(x(1 + logx)^2)` = `log2/(1 + log2)`.

Explanation:

Let I = `int_1^3 ("d"x)/(x(1 + logx)^2)` 

Put 1 + log x = t ⇒ `1/x  "d"x` = dt

∴ I = `int_1^(1 + log2) "dt"/"t"^2 = [- 1/"t"]_1^(1 + log 2)`

= `- 1/(1 + log 2) + 1`

= `(1 + log2 - 1)/(1 + log 2)`

∴ I = `log2/(1 + log2)`