My Math Forum integration by parts xlny

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February 24th, 2018, 04:45 PM   #11
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Quote:
 Originally Posted by greg1313 $$\int y\ln y\,dy=y(y\ln y-y)-\int y\ln y-y\,dy=y(y\ln y-y)-\int y\ln y\,dy+\int y\,dy$$
$$u=y$$

$$dv=\ln y$$

Ok?

 February 24th, 2018, 05:39 PM #12 Senior Member   Joined: Apr 2017 From: New York Posts: 155 Thanks: 6 if u=y and dv= lny then du=dy and v=ylny-y perfect. thanks a lot sir. I couldn't figure out at first how v was now very clear. thanks million. Thanks from greg1313
 February 24th, 2018, 06:17 PM #13 Senior Member   Joined: Apr 2017 From: New York Posts: 155 Thanks: 6 From that point I will use another version as u=lny and dv=y .... do you recommend that path or ? I know you used u=y and dv=lny, but I was not practical integrating lny and couldn't see the result fast. Last edited by skipjack; February 24th, 2018 at 08:20 PM.
 February 24th, 2018, 08:18 PM #14 Global Moderator   Joined: Dec 2006 Posts: 20,978 Thanks: 2229 Let $u = \ln(1 + x)$ and $v = \frac12\left(x^2 - 1\right)$, so that $\displaystyle du = {\small\frac{1}{1 + x}}\,dx$ and $dv = x\, dx$, then \displaystyle \begin{align*}\!\int\! x\ln(1 + x)\,dx &= \frac12\left(x^2- 1\right)\ln(1 + x) - \frac12\!\int\! \frac{x^2 - 1}{1 + x}\, dx \\ &= \frac12\left(x^2 - 1\right)\ln(1 + x) - \frac12\!\int (x - 1)\, dx \\ &= \frac12(x^2-1)\ln(1 + x) - \frac14(x - 1)^2 + \text{C} \\ \end{align*} Thanks from greg1313

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