Problem:
Evaluate$$\int_0^1 \dfrac{\ln \left(a+\sqrt{a^2+1}\right)}{a\sqrt{a^2+1}}da$$
It was suggested that I try Integration By Parts.
Problem:
Evaluate$$\int_0^1 \dfrac{\ln \left(a+\sqrt{a^2+1}\right)}{a\sqrt{a^2+1}}da$$
It was suggested that I try Integration By Parts.
If we use the substitution $a=\sinh u$, then $u=\log v$, we are left with:
$$ I = \int_{0}^{\log(1+\sqrt{2})}\frac{u\,du}{\sinh u}=\int_{1}^{1+\sqrt{2}}\frac{2\log v}{v^2-1}\,dv = \left.\frac{d}{d\alpha}\int_{1}^{1+\sqrt{2}}\frac{v^{\alpha}\,dv}{v^2-1}\right|_{\alpha=0}$$ that ultimately depends on $\frac{\pi^2}{4}$, the product of some logarithms, $\text{Li}_2(1-\sqrt{2})$ and $\text{Li}_2(\sqrt{2}-1)$:
$$ I = \frac{\pi^2}{4}+\log(\sqrt{2}+1)\log(\sqrt{2}-1)+\text{Li}_2(1-\sqrt{2})-\text{Li}_2(\sqrt{2}-1),$$
since: $$ \int\frac{\log v}{v+1}\,dv=\log v \log(1+v)+\text{Li}_2(-v), $$ $$ \int\frac{\log v}{v-1}\,dv = -\text{Li}_2(1-v). $$ Not exactly a trivial integral. The closed form:
$$ I = \frac{\log^2(1+\sqrt{2})}{2} $$
follows for the functional identity for the dilogarithm: $$ \text{Li}_2(1-x)+\text{Li}_2(1-x^{-1})=-\frac{1}{2}\log^2 x $$
that is straightforward to prove through differentiation.