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$\newcommand{\d}[1]{\mathop{\mathrm d #1}}$ To integrate $x\ln (x)$, let $u = x$ and $\d v = \ln(x)\d x$. It becomes

$$ \int x\ln(x)\d x = x\int \ln(x)\d x\,-\int \int \ln(x) \d {x^2} $$

simplified to

$$ \int x\ln(x)\d x = x(x\ln(x) -x)\,-\int(x\ln(x) - x)\d x $$

break up integral and solve integral $x \d x$

$$ \int x\ln(x)\d x = x(x\ln(x) -x)\,-\int x\ln(x) \d x - \frac{1}{2}x^2 \qquad(\star)$$

add (integral $x \ln x \d x$) to both sides then divide by two

$$ \int x\ln(x)dx = \frac{x(x\ln(x) -x) - \frac{1}{2}x^2}{2}$$

and this is apparently not the answer. The answer is actually

$$\frac{1}{4}x^2(-1+2\log (x))$$

which is not the same function.

What am I doing wrong?

I can do this problem fine if I assign $u$ and $v$ the other way around, but this should still work and I feel like I'm making a dumb mistake somewhere.

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  • $\begingroup$ Hint: take $dv = 1$ and $u = x\ln x$. It's a lot easier. $\endgroup$
    – Deepak
    Oct 2, 2016 at 7:43
  • $\begingroup$ You dropped a minus in the step from $\int x\ln x -x\mathrm{d}x=\int x\ln x \mathrm{d}x-x^2/2$ $\endgroup$
    – Jam
    Oct 2, 2016 at 7:43
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    $\begingroup$ $\uparrow$ marked with a star $\endgroup$ Oct 2, 2016 at 7:45

3 Answers 3

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$$\begin{cases} du=xdx \\ v=\ln { x } \end{cases}\Rightarrow \begin{cases} u=\frac { { x }^{ 2 } }{ 2 } \\ dv=\frac { dx }{ x } \end{cases}\\ \int { x\ln { xdx } } =\frac { { x }^{ 2 }\ln { x } }{ 2 } -\int { \frac { x }{ 2 } dx= } \frac { { x }^{ 2 }\ln { x } }{ 2 } -\frac { { x }^{ 2 } }{ 4 } +C$$

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$$\begin{align}∫x\ln x \,\mathrm{d}x&=x(x\ln x−x)−∫(x\ln x−x)\,\mathrm{d}x \\&=x(x\ln x−x)+∫(-x\ln x+x)\,\mathrm{d}x \\&=x(x\ln x−x)+∫(-x\ln x)\,\mathrm{d}x+\frac{1}{2}x^2 \end{align}$$

It's quite easy to drop a minus in algebra like this (and everyone does) so I often use brackets in my integrals. Just keep at it and you'll learn to stop making these mistakes.

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  • $\begingroup$ Oh thank you. Looks like I was just being a retard. $\endgroup$ Oct 3, 2016 at 0:39
  • $\begingroup$ @user2958456 No worries man $\endgroup$
    – Jam
    Oct 3, 2016 at 2:09
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Here is just a different approach. Not necessarily better, just different: Why not u-sub $lnx=t$ so that $dx=e^tdt$. The integral becomes then $\int{te^{2t}dt}$. This can be done MUCH easier now with integration by parts. You can even go one step further: Set $2t=v$ and then $dt=0.5dv$. Then the integral becomes $\frac{1}{4}\int{ve^vdv}$. Why would you do that? Because now the original integral has been completely reduced to a Standard integral. Of course, this is also done through integration by parts but more basic than this is not possible. Problems like $\int{xe^xdx}$ can be found in any standard Calculus book. All you need to do is back subs, which is standard algebra. I am a big proponent for transforming given integrals in terms of standard integrals because the work becomes so much easier and if you have once set up a list of standard integrals, you don't have to integrate over and over again...

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