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$$\int_{-\infty}^{\infty}\frac{\ln(x^4 + 1)}{x^2 + 1}dx$$ I tried Feynman's technique but got the wrong answer. I also tried by factoring $x^4+1$ as $(x^2+i)(x^2-i)$ where $i=\sqrt{-1}.$

Anne Bauval
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jayesh kumar gupta
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    This is a duplicate, almost certainly. – FShrike Jun 25 '23 at 16:13
  • Hi, welcome to Math SE. I agree with @FShrike, but if you know the residue theorem here's a hint: write the integral as$$2\Re\int_{-\infty}^\infty\frac{\ln(x+e^{i\pi/4})(x+e^{3i\pi/4})}{x^2+1}dx$$then use a semicircular contour in the upper half-plane. You should get $2\pi\ln(2+\sqrt{2})$. – J.G. Jun 25 '23 at 16:14
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    thx but is there any other way to integrate without using contour integration as i havent properly studied it – jayesh kumar gupta Jun 25 '23 at 16:22

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