Question

`int_e^(e^2)[1/(logx) - 1/(logx)^2]dx` = ______

Options

• `(e - 2)/2`

• `(e(e - 2))/2`

• `e^2/2 - 2e`

• `e^2/2 + 2e`

Answer 1

`int_e^(e^2)[1/(logx) - 1/(logx)^2]dx = underline((e(e - 2))/2)`

Explanation: 

Let I = `int_e^(e^2)[1/(logx) - 1/(logx)^2]dx` 

Put log x = t ⇒ x = e^t ⇒ dx = e^t dt

∴ I = `int_1^2e^t(1/t - 1/t^2)dt`

= `[e^t/t]_1^2` ................[∵ ∫e^x[f(x) + f'(x)]dx = e^xf(x) + c]

= `e^2/2 - e`

∴ I = `(e(e - 2))/2`