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February 24th, 2018, 01:03 PM   #1
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integration by parts xlny

Hi guys,
Where do I make mistake.
I have rewritten three times still getting the same wrong answer
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February 24th, 2018, 01:34 PM   #2
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Your attachment is difficult to read. Type it in directly.
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February 24th, 2018, 02:16 PM   #3
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Line 6: $-\frac12y$ should be $+\frac12y$

But that probably doesn't matter as the integral in line 4 is wrong. It should be $\int y^2 + 2y \,\mathrm dy$.

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February 24th, 2018, 02:34 PM   #4
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ok thanks checking quickly again.

sorry for small picture. I am attaching better one ( I guess )
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February 24th, 2018, 02:42 PM   #5
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v times du = 1/2(y-1)
is that wrong? on line for to set up integral.
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February 24th, 2018, 02:42 PM   #6
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on line four*
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February 24th, 2018, 02:44 PM   #7
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ok never mind yes line four is wrong
Let me work on that again and send
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February 24th, 2018, 02:55 PM   #8
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How is it now, please?
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February 24th, 2018, 03:50 PM   #9
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$$y=1+x$$

$$dy\,=\,dx$$

$$\int(y-1)\ln y\,dy=\int y\ln y\,dy-\int\ln y\,dy$$

$$\int\ln y\,dy=y\ln y-\int\,dy=y\ln y-y+C$$

$$\int y\ln y\,dy=y(y\ln y-y)-\int y\ln y-y\,dy=y(y\ln y-y)-\int y\ln y\,dy+\int y\,dy$$

$$2\int y\ln y\,dy=y(y\ln y-y)+\frac{y^2}{2}+C$$

I'll let you finish up...
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February 24th, 2018, 04:42 PM   #10
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Up to here I understood perfectly and liked your method very much indeed.

could you explain what is u, and dv at this line:

∫ylnydy=y(ylny−y)−∫ylny−ydy=y(ylny−y) ∫ylnydy+∫ydy

I did not understand this part

and I know in the last line I will substitute y=1+x but I didn't want to do by heart just wanted to learn
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