Integrate $\int_0^1\frac{\text{sech}^{-1}(x)(1-2x^2)}{1+4x^4}dx.$

Question

I'm having problem to evaluate this integral.

$$\int_0^1\frac{\text{sech}^{-1}(x)(1-2x^2)}{1+4x^4}dx.$$

Where should I start?

I know that $\text{sech}^{-1}(x)=\ln \left(\frac{1+\sqrt{1-x^2}}{x}\right )$.

Thanks.

Answer 1

The change of variables $s = \text{sech}(x)$ makes this into

$$ \int_0^\infty \frac{s (\cosh(s)^2 - 2) \sinh(s)}{\cosh(s)^4+4}\; ds $$

Now $\cosh(s) > 1$, and for $c > 1$ we have $$ \frac{c^2-2}{c^4+4} = \sum_{k=1}^\infty (\cos(k\pi/2) + \sin(k\pi/2)) 2^{k-1}/c^{2k} $$ It seems to me that $$ \int_0^\infty \frac{s \sinh(s)}{\cosh(s)^{2k}}\; ds = \frac{2^{1-2k} \pi (2k-2)!}{(k-1)!^2 (2k-1)}$$

so your integral becomes

$$ \sum_{k=1}^\infty (\cos(k\pi/2) + \sin(k\pi/2)) \frac{2^{-k}\pi (2k-2)!}{(k-1)!^2(2k-1)} = \frac{\pi}{2} \arcsin \left(\frac{\sqrt{5}-1}{2}\right) $$