Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We have to integrate $$\int\frac{2x \ln x}{\sqrt{(x^2-9)}} \mathrm dx$$

Is it right to use Integration by Parts?

I tried to substitute it with $$u = \log x,\: \mathrm du = \frac 1x \mathrm dx;$$ $$v = x^2 - 9,\: \mathrm dv = 2x \mathrm dx.$$

But then I'm stuck with substituting it within the original equation because from $\mathrm du = \dfrac 1x \mathrm dx,$ and $\mathrm dv = 2x \mathrm dx,$ there will be two $\mathrm dx$'s to substitute and from $\mathrm du = \dfrac 1x \mathrm dx,$ the $x$ will go to the denominator and I don't know what to do any more.

share|cite|improve this question

You need to make the integral into only two parts, $u$ and $dv.$ So you should let $u= \ln x$ and $dv = \dfrac{2x}{\sqrt{x^2-9}} dx.$

Then using those choices, work out what $du$ and $v$ must be, and put it all in the formula $\displaystyle \int u dv = uv - \int v du.$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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