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 Calculus Calculus Math Forum

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 Tags integration, parts, xlny Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Leonardox Calculus 4 February 22nd, 2018 04:18 PM aaronmooy47 Calculus 6 April 2nd, 2017 01:33 PM fredlo2010 Calculus 8 September 2nd, 2014 12:49 PM Zynoakib Calculus 2 March 14th, 2014 05:21 PM tnutty Calculus 1 February 23rd, 2009 01:31 PM

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