Why integration by part (not partial) is considered everywhere useful?

Integration by parts is:

$$\int\color{black}{u(x)}\,\color{black}{v'(x)}\,\mathrm dx=\color{black}{u(x)}\color{black}{v(x)}-\int\color{black}{u'(x)}\color{black}{v(x)}\,\mathrm dx$$

But for example if I have function $f(x)=\ln(x)$ how can I find its integral using integration by parts? Otherwise integration by parts is not everywhere useful

-
What do you mean by finding an integral using the chain rule? –  Michael Albanese Dec 27 '13 at 14:06
What is that of being "everywhere useful"!? –  JMCF125 Dec 27 '13 at 14:10

Hint:

$$\int\color{blue}{u(x)}\,\color{red}{v'(x)}\,\mathrm dx=\color{blue}{u(x)}\,\color{red}{v(x)}-\int\color{blue}{u'(x)}\,\color{red}{v(x)}\,\mathrm dx$$

$$\int\color{blue}{\ln(x)}\,\mathrm dx=\int\color{blue}{\ln(x)}\cdot\color{red}1\,\mathrm dx=\,\,?$$

-

Not sure why you would expect a particular computational method to be useful for every problem. Try $u(x)=\ln x$ and $v(x)=x$.

-
Minor typo: you probably mean $v(x)=1$, not $x$. –  Ian Mateus Dec 27 '13 at 14:09
@IanMateus No, Mark is correct: $\frac{\mathrm d}{\mathrm dx}x=1$. That's why $v'(x)=1$ and $v(x)=x$ and not the other way around. –  user93957 Dec 27 '13 at 14:14
I got it backwards! Thanks for the correction :-) –  Ian Mateus Dec 27 '13 at 14:19
@IanMateus You're welcome! ;-) –  user93957 Dec 27 '13 at 14:19