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Today one of my friends asked me what's the $$\int \log x \,\, \mathrm{d}x$$ and I was unable to answer. I think we just take the Taylor series and integrate it . Is that all or something else.

Note: I am grade $11$ student so I don't know any advanced calculus and all hyperbolic functions.

Jeel Shah
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Archis Welankar
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    Write your integral as $\int 1. \ln x dx$ and apply integration by parts. –  Mar 02 '16 at 12:45
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    Oh so its $xlogx-x$ – Archis Welankar Mar 02 '16 at 12:47
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    $$\displaystyle\bbox[1ex, border:solid 1.5pt #e10000]{\int \log x ,dx = x\log x - x + c} \impliedby \dfrac{d}{dx}\big(x\log x - x + c\big) = \log x + \dfrac{x}{x} - 1 = \log x$$ – Vlad Mar 02 '16 at 12:47
  • A grade 11 student who first thinks of Taylor series and term-by-term integration instead of integration by parts. Well that's new. – Trogdor Mar 02 '16 at 13:12
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    Im new to whole integration and being here for a while i saw many people using the word taylor series so i thought so – Archis Welankar Mar 02 '16 at 13:19
  • The spoiler here (that Parts should be used) is that $\log x$ has a simple derivative. – John Joy Mar 02 '16 at 16:01

2 Answers2

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Partial integration does help here. We have \begin{align*} \int \log x \, dx &= \int 1 \cdot \log x \, dx \\ &= x \log x - \int x \cdot \frac 1x \, dx \\ &= x \log x - x+C. \end{align*}

zz20s
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martini
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A little longer is with a change of variable:

$$\int\ln x\mathrm d x\to\begin{matrix} x=e^y\\\mathrm d x=\mathrm d e^y=e^y\mathrm d y\end{matrix}\to\int ye^y\mathrm d y=ye^y-e^y+C=x(\ln x-1)+C$$

Of course this answer is "worst" than a direct manipulation as other answer shows but I want just to illustrate a "different" way to do it.

Masacroso
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