I-1-lnx-dx-

Question Number 172839 by DAVONG last updated on 02/Jul/22

I = \int \frac{1}{lnx} dx = ?

Answered by puissant last updated on 02/Jul/22

I = \int \frac{1}{l n x} d x; x = e^{u} \Rightarrow d x = e^{u} d u

I = \int \frac{e^{u}}{u} d u = \int \sum_{n ⩾ 0} \frac{u^{n - 1}}{n!} d u

= \int (\frac{1}{u} + \sum_{n ⩾ 1} \frac{u^{n - 1}}{n!}) d u

= l n u + \sum_{n ⩾ 1} \frac{u^{n}}{n . n!} + C

= l n (l n x) + \sum_{n ⩾ 1} \frac{{l n}^{n} x}{n . n!} + C

\dots \dots . L e p u i s s a n t \dots \dots

Commented by ghakhan last updated on 03/Jul/22

c a n y o u e x p l a i n h o w d i d g e t \frac{1}{u} i n t h e t h i r d s t e p p l e a s e ?

Commented by puissant last updated on 03/Jul/22

\sum_{n ⩾ 0} \frac{u^{n - 1}}{n!} = \frac{u^{0 - 1}}{0!} + \sum_{n ⩾ 1} \frac{u^{n - 1}}{n!}

= \frac{1}{u} + \sum_{n ⩾ 1} \frac{u^{n - 1}}{n!} .

Commented by DAVONG last updated on 03/Jul/22

Thanks teacher!

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