Question

integration of 1/ log x
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Answer 1

Answer:1/log x) dx = t ^−1 . e^t + t ^−2. e^t + 2 . t ^−3 .e^t+ 6.t ^−4 . e^t + …..

Proof :

Let t = logx => e^t = x

=> dt= (1/x) . dx or dx = e^t . dt

∴, ∫(1/log x) dx = ∫e^t. dt/t = ∫ t^-1 . e^t. dt

Using integration by parts,

∫udv = uv - ∫vdu

In ∫t^-1 . e^t. dt

Take, u=t ^−1 => du=−t^−2.dt

∫dv=∫e^tdt => v=e^t

Now substituting,

∫(1/log x) dx = t^−1.e^t−∫e^t (−t^−2) dt

Similarly using integration by parts, we have to integrate ∫t^−2.e^t. dt

This is a never ending integral and the approximate value is:

∫(1/log x) dx = t ^−1 . e^t+t ^−2. e^t+ 2 . t ^−3 .e^t+ 6.t ^−4 . e^t + …..

Hope it helps!!

Cheers,

Andy

Step-by-step explanation: