tebollahb

2022-01-21

Is there a different approach to evaluate $\int \mathrm{ln}\left(x\right)dx$?

Melinda McCombs

Expert

$\mathrm{ln}x=\sum _{n=1}^{\mathrm{\infty }}{\left(-1\right)}^{n+1}\frac{1}{n}{\left(x-1\right)}^{n}$
$\int \mathrm{ln}xdx=\sum _{n=1}^{\mathrm{\infty }}{\left(-1\right)}^{n+1}\frac{1}{n}\cdot \frac{1}{1+n}{\left(x-1\right)}^{n+1}$
$=\sum _{n=1}^{\mathrm{\infty }}{\left(-1\right)}^{n+1}\left(\frac{1}{n}-\frac{1}{n+1}\right){\left(x-1\right)}^{n+1}$
$=\left(x-1\right)\sum _{n=1}^{\mathrm{\infty }}{\left(-1\right)}^{n+1}\frac{1}{n}{\left(x-1\right)}^{n}+\sum _{n=1}^{\mathrm{\infty }}{\left(-1\right)}^{n+2}\frac{1}{n+1}{\left(x-1\right)}^{n+1}$
$=\left(x-1\right)\mathrm{ln}x+\mathrm{ln}x-\left(x-1\right)$
$=x\mathrm{ln}x-x+C$

Expert

Im

RizerMix

Expert

For fun, a bit of trickery: Let $x={e}^{y}$. Then: $\int \mathrm{ln}xdx=\int y{e}^{y}dy;$ Now look at : $\frac{d}{dy}\left(y{e}^{y}\right)={e}^{y}+y{e}^{y}.$ Thus: $\int y{e}^{y}dy=$ $\int \frac{d}{dy}\left(y{e}^{y}\right)-\int {e}^{y}dy=$ $y{e}^{y}-{e}^{y}=x\mathrm{ln}x-x+C.$