Question

Integrate the following w.r.t.x : log (log x)+(log x)^–2 

Answer 1

Let I = `int [log (logx) + (logx)^-2]*dx`

= `int [log(logx) + 1/(log x)^2]*dx`
Put log x = t
∴ x = e^t
∴ x = e^t·dt

∴ I = `int (log t + 1/t^2)e^t*dt`

= `int e^t (log t + 1/t - 1/t + 1/t^2)*dt`

= `int [e^t (log t 1/t) + e^t (-1/t + 1/t^2)]*dt`

= `inte^t (log t + 1/t)*dt - int e^t (1/t - 1/t^2)*dt`

= I₁ – I₂ 

In I₁, Put f(t) = log t. Then f'(t) = `(1/t)`

∴ I₁ = `int e^t [f(t) + f'(t)]*dt`

= `e^t f(t)`

= `e^t log t`

In I₂, Put g(t) = `(1/t)`. Then g'(t) = `-(1/t^2)`

∴ I₂ = `int e^t ["g"(t) + "g"'(t)]*dt`

= `e^t "g" (t)`

= `e^t*(1/t)`

∴ I = `e^t log t - e^t/t + c`

= `xlog (logx) - x/logx + c`.