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Said integral being $$\int\frac1x\ln(\frac1{x^2+1})dx$$ I've tried solving it with normal integration by parts exploiting that $$\ln(\frac1{x^2+1})=(-1)\ln(x^2+1)$$ but no dice. I'm trying to find a way to shoehorn in $arctanx$ as a primitive of $\frac1{x^2+1}$ without making the rest unreasonably complicated.

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  • $\begingroup$ Do you expect to find a closed-form, elementary anti-derivative? $\endgroup$ – StackTD Jul 12 '18 at 11:49
  • $\begingroup$ I'm actually just supposed to find the definite integral for 1<x<2. I tried to pose y=(starting integral) and find it indirectly but it just loops on itself with no usable coefficents. $\endgroup$ – Moss Jul 12 '18 at 11:51
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    $\begingroup$ It's always useful to add this information and/or relevant context; sometimes you can find a definite integral without explicitly finding an anti-derivative. $\endgroup$ – StackTD Jul 12 '18 at 11:53
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HINT: $$\ln\left(\frac1{x^2+1}\right) = -\ln \left(x^2+1\right)$$ $$-\int\frac{\ln \left(x^2+1\right)}{x}dx$$ Set $$u = -x^2 \implies dx = -\frac{1}{2x}\,du$$ $$-\frac{1}{2}\int\frac{\ln\left(1-u\right)}{u}du$$

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  • $\begingroup$ I've tried to put the -1/2 aside and resolve the new integral by parts but I'm still stumped. $\endgroup$ – Moss Jul 12 '18 at 12:14
  • $\begingroup$ it's because it's not a standard integral.See: en.wikipedia.org/wiki/Polylogarithm $\endgroup$ – Darío A. Gutiérrez Jul 12 '18 at 12:18

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