How would you solve $\displaystyle \int \ln (\sqrt{x^2+x+1}) dx$? I don't see any good substitution nor a good integration by part. I'm not asking the whole solution, just the way you'd solve it. Thanks.
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Note that $\displaystyle \log((x^2+x+1)^{1/2}) = \frac{1}{2}\log\Big(\frac{x^3-1}{x-1}\Big)$ $\displaystyle =\frac{1}{2}\Big(\log(x^3-1)-\log(x-1)\Big)$
$\displaystyle \frac{1}{2}\int1.\ln(x^2+x+1)dx=\frac{x}{2}\ln(x^2 +x+1)+$...............
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