I came across while solving some problem that I need the exact/ approximate value of following integral. I need an upper bound of $-0.4$ or $-0.5$. Is there some way or programming tools.
$$\int_0^{\sqrt{2}} \ln\left(\dfrac{x^6+8x^4+16x^2+4}{x^6+8x^4+15x^2+6}\right)\,\mathrm dx$$
This is not homework problem. Just while solving some calculus problem, it came out
On
Ask Wolfram alpha or a computer algebra package at the university...
It gives $-0.211212$ as an approximation.
Write numerator and denominator as $$\dfrac{x^6+8x^4+16x^2+4}{x^6+8x^4+15x^2+6}=\frac{(x^2+a)(x^2+b)(x^2+c)}{(x^2+d)(x^2+e)(x^2+f)}$$ where $a,b,c,d,e,f$ are all positive numbers.
Expand the logarithm to face $$I(p)=\int_0^{\sqrt 2} \log(x^2+p)=\sqrt{2} (\log (p+2)-2)+2 \sqrt{p} \tan ^{-1}\left(\frac{\sqrt{2}}{\sqrt{p}}\right)$$
If I am not mistaken, $a,b,c$ are given by $$\frac{8}{3}-\frac{8}{3} \cos \left(\frac{2 \pi k}{3}-\frac{1}{3} \cos ^{-1}\left(\frac{5}{32}\right)\right)\qquad \text{with} \qquad k=0,1,2$$ and $c,d,e$ by $$\frac{8}{3}-\frac{2}{3} \sqrt{19} \cos \left(\frac{2 \pi k}{3}-\frac{1}{3} \cos ^{-1}\left(-\frac{53}{19 \sqrt{19}}\right)\right)\qquad \text{with} \qquad k=0,1,2$$
Edit
If we just want an approximate value of the integral, using the simplest Padé approximant built around $x=1$ (more or less the midpoint of the integration range), we have $$\dfrac{x^6+8x^4+16x^2+4}{x^6+8x^4+15x^2+6}\sim \frac{\frac{3389 (x-1)^2}{4520}+\frac{4691 (x-1)}{3390}+\frac{29}{30}}{\frac{1799 (x-1)^2}{2260}+\frac{2177 (x-1)}{1695}+1}$$
Thus, completing the squares, we have $$\log\left(\dfrac{x^6+8x^4+16x^2+4}{x^6+8x^4+15x^2+6}\right)\,dx \sim\log\left(\dfrac{a (x-b)^2+c}{d (x-e)^2+f}\right)\,dx$$ and $$J=\int \log \left(a (x-b)^2+c\right)\,dx$$ is given by $$J=(x-b) \log \left(a (x-b)^2+c\right)+\frac{2 \sqrt{c} \tan ^{-1}\left(\frac{\sqrt{a} }{\sqrt{c}}(x-b)\right)}{\sqrt{a}}-2 (x-b)$$ This should lead to a value equal to $-0.212276$ to be compared to the exact $−0.211212$ given by Wolfram Alpha (this makes a relative error of $0.5$%.