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February 24th, 2018, 05:45 PM   #11
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$$\int y\ln y\,dy=y(y\ln y-y)-\int y\ln y-y\,dy=y(y\ln y-y)-\int y\ln y\,dy+\int y\,dy$$
$$u=y$$

$$dv=\ln y$$

Ok?
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February 24th, 2018, 06:39 PM   #12
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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.
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February 24th, 2018, 07:17 PM   #13
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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 09:20 PM.
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February 24th, 2018, 09:18 PM   #14
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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
$\displaystyle \begin{align*}\!\int\! x\ln(1 + x)\,dx &= \frac12\left(x^2- 1\right)\ln(1 + x) - \frac12\!\int\! \frac{x^2 - 1}{1 + x}\, dx \\
&= \frac12\left(x^2 - 1\right)\ln(1 + x) - \frac12\!\int (x - 1)\, dx \\
&= \frac12(x^2-1)\ln(1 + x) - \frac14(x - 1)^2 + \text{C} \\
\end{align*}$
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