Integral $I = \int_0^1 \frac{\sqrt{x^3 + 1}}{(x^5 + x^2 + 1)} \ln\left(\frac{x^6 + x^3 + 1}{x^6 – x^3 + 1}\right) \, dx.$

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$$I = \int_0^1 \frac{\sqrt{x^3 + 1}}{(x^5 + x^2 + 1)} \ln\left(\frac{x^6 + x^3 + 1}{x^6 – x^3 + 1}\right) \, dx.$$

I graphed the integral in Wolfram Alpha and got this result:

1

I need some help in the first steps of solving this integral. Thanks.

Best Answer

$$I = \int_0^1 \frac{\sqrt{x^3 + 1}}{(x^5 + x^2 + 1)} \ln\left(\frac{x^6 + x^3 + 1}{x^6 - x^3 + 1}\right) \, dx.$$ WolframAlpha gives this approximate :

enter image description here $$I\simeq 0.252034156031883...$$ The ISC (Inverse Symbolic Calculator) from CECM (http://wayback.cecm.sfu.ca/projects/ISC/ISCmain.html) doesn't find a closed form but gives the next analytic formulas which numerical approximates have 8 correct first digits:

enter image description here

My own ISC (https://fr.scribd.com/doc/14161596/Mathematiques-experimentales) also doesn't find a closed form but gives a fanciful formula which numerical approximate has 10 correct first digits: $$I\simeq \sqrt{\frac{\exp{(-e)}}{\sin(1)}}-\sin\left(\frac{\exp(-\pi)}{\cosh(1)} \right)\simeq 0.252034156023...$$ If the ISCs had found a possible (but not proved) closed form the number of correct digits would be much larger.

Conclusion : All this draw us to think that the integral cannot be expressed on a closed form made of a limited number of standard functions. One have to be satisfied with numerical approximate.

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