Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

closed as unclear what you're asking by Jonas Meyer, Ittay Weiss, Grigory M, dustin, Did Jan 18 at 19:03

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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

2 Answers 2

up vote 7 down vote accepted


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

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

share|improve this answer

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

share|improve this answer
Minor typo: you probably mean $v(x)=1$, not $x$. –  Ian 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 Dec 27 '13 at 14:19
@IanMateus You're welcome! ;-) –  user93957 Dec 27 '13 at 14:19

Not the answer you're looking for? Browse other questions tagged or ask your own question.