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I was recently doing some exercises on Integrals. I already solved the same integral but with $x^2 + 1$ as well as $x^4 + 1$ as Argument of the logarithm. I’m struggling with this case though and WolframAlpha fails to present a step-by-step solution, can you guys help me out on this one?

Giorgos Giapitzakis
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Mathisnothealthyforu
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    Hint: if $\Im w>0$ a semicircular contour in the upper half plane gives $\int_{-\infty}^\infty\frac{\ln(x+w)}{x^2+1}dx=\pi\ln(i+w)$ so$$\int_{-\infty}^\infty\frac{\ln\left(x^2+\left(2\Re w\right)x+|w|^2\right)}{x^2+1}dx=\pi\ln\left(|w|^2+1+2\Im w\right).$$Now, using sixth roots of $-1$, express $x^6+1$ as the product of three quadratics with real coefficients, including $x^2+1$, which you've solved already. – J.G. Jan 11 '23 at 23:50
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    See https://math.stackexchange.com/a/4515389/686284 – Quanto Jan 12 '23 at 01:32
  • Thank you all!! – Mathisnothealthyforu Jan 12 '23 at 08:23

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$$\int_0^1\frac{\ln(1+x^6)}{1+x^2}dx= \underbrace{\int_{0}^{\infty} \frac{\ln \left(x^{2}+1\right)}{1+x^{2}} d x}_{=\pi \ln 2}+ \underbrace{\int_{0}^{\infty} \frac{\ln \left(x^{4}-x^{2}+1\right)}{1+x^{2}} d x }_{K} $$

For integral $K$, we use the formula founded in my post stating that $$J(a)=\int_{0}^{\infty} \frac{\ln \left(1+2 x \sin a+x^{2}\right)}{1+x^{2}} d x =\pi\ln \left|2 \cos \frac{a}{2}\right|+|a| \ln \left|\tan \frac{a}{2}\right|-2 \operatorname{sgn} (a) \int_{0}^{\frac{|a|}{2}} \ln (\tan x) d x$$ Then $$ \begin{aligned} K &=\int_{0}^{\infty} \frac{\ln \left(x^{4}-x^{2}+1\right)}{1+x^{2}} d x \\ &=\int_{0}^{\infty} \frac{\ln \left(x^{2}+\sqrt{3} x+1\right)}{1+x^{2}} d x+\int_{0}^{\infty} \frac{\ln \left(x^{2}-\sqrt{3} x+1 \right) }{1+x^{2}} dx\\ &=J\left(\frac{\pi}{3}\right)+J\left(-\frac{\pi}{3}\right) \\ &=2 \pi \ln \left(2 \cos \frac{\pi}{6}\right) \\ &=\pi \ln 3 \end{aligned} $$

$$ \boxed{ \int_{0}^{\infty} \frac{\ln \left(1+x^{6}\right)}{1+x^{2}} d x=\pi \ln 2+\pi \ln 3=\pi \ln 6 } $$

Lai
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