Several variations of Ahmed integrals: $\int_0^1\frac1{(1+y^2)\sqrt{2+y^2}}\arcsin\left(\sqrt{\frac{2-y^2}{4}}\right )\text{d}y=\frac{11\pi^2}{288}.$

Question

The following are four integrals resembling Ahmed integrals: $$ \begin{aligned} &I_1=\int_{0}^{1} \frac{1}{\left ( 1+y^2 \right )\sqrt{2+y^2} } \arcsin\left ( \sqrt{\frac{2-y^2}{4} } \right ) \text{d}y = \frac{11\pi^2}{288},\\ &I_2=\int_{0}^{1} \frac{1}{\left ( 1+y^2 \right )\sqrt{2+y^2} } \arcsin\left ( \sqrt{\frac{1-y^2}{3} } \right ) \text{d}y = \frac{\pi^2}{8}-\frac{\pi}{2} \arcsin\left ( \frac{1}{\sqrt{3}} \right ),\\ &I_3=\int_{0}^{1} \frac{1}{\left ( 1+y^2 \right )\sqrt{2+y^2} } \arctan\left ( \frac{\sqrt{(1-y^2 )\left ( 2+y^2 \right ) }} {2}\right ) \text{d}y =\pi\arctan\left ( \frac{1}{\sqrt{2} } \right ) -\frac{\pi^2}{6},\\ &I_4=\int_{0}^{1} \frac{1}{\left ( 1+y^2 \right )\sqrt{2+y^2} } \arctan\left ( {\sqrt{\frac{4-y^4}5}}\right ) \text{d}y =\frac\pi2\arctan\left ( \sqrt{\frac35} \right ) -\frac{\pi^2}{15}.\\ \end{aligned} $$ I derived them by some complex tricks, while two alternative approaches sprung up into my mind. We can handle them either by $S(\alpha,\beta,\gamma)$ or @Zacky's approach. For instance, using the latter we obtain an equivalent of $I_2$: $$ \int_{0}^{\infty} \frac{\left ( x^4-2x^2+1 \right ) }{x(1+x^2)(1+2x^2)(1+3x^2)} \ln\left ( \left | \frac{1-x}{1+x} \right | \right ) \text{d}x =\pi\arcsin\left ( \frac{1}{\sqrt{3} } \right ) -\frac{\pi^2}{12} \sim1.1111194977. $$ This monster is actually solvable for the friendly interval, and we may employ residues to have a rapid verfication. Moreover, by borrowing some ideas from this previously posted question, we have $$ \small\int_{0}^{1} \int_{0}^{1} \frac{1}{\left ( 1+x^2 \right )\left ( 1+y^2 \right ) \sqrt{3+x^2+y^2} }\arcsin\left ( \sqrt{\frac{3+x^2+y^2}{5} } \right ) \text{d}x\text{d}y =2\pi\int_{0}^{1} \frac{1}{\left ( 1+x^2 \right ) \sqrt{2+x^2} }\arcsin\left ( \sqrt{\frac{2+x^2}{5} } \right ) \text{d}x -\frac{\pi^2}{2}\arcsin\left ( \frac{1}{\sqrt{5} } \right ) +\frac{\pi^3}{160}, $$ $$ \small\int_{0}^{1} \int_{0}^{1} \frac{1}{\left ( 1+x^2 \right )\left ( 1+y^2 \right ) \sqrt{3+x^2+y^2} }\arcsin\left ( \sqrt{\frac{3+x^2+y^2}{6} } \right ) \text{d}x\text{d}y =\pi\int_{0}^{1} \frac{1}{\left ( 1+x^2 \right ) \sqrt{2+x^2} }\arcsin\left ( \sqrt{\frac{2+x^2}{6} } \right ) \text{d}x -\frac{\pi^2}{5}\arcsin\left ( \frac{1}{\sqrt{6} } \right ) +\frac{\pi^3}{960}. $$

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Questions.

1. Do we have any alternative proofs and new results of such integrals? Especially with factors $\frac{1}{(1+y^2)\sqrt{2+y^2}}$.
2. On whether it is possible to cope with the multiple integration?

Answer 1

Ahmed integral: $\int_0^1 \frac{\tan^{-1}\sqrt{2+x^2}}{(1+x^2)\sqrt{2+x^2}}\ dx= \frac{5\pi^2}{96}$.

\begin{aligned} I_1=&\int_{0}^{1} \frac{1}{\left ( 1+y^2 \right )\sqrt{2+y^2} } \sin^{-1}\sqrt{\frac{2-y^2}{4} } \ \text{d}y \\ \overset{ibp}=&\ \frac{\pi^2}{36}+\int_0^1 \frac{y\tan^{-1}\frac y{\sqrt{2+y^2}}}{\sqrt{4-y^4}}dy\\ =&\ \frac{\pi^2}{36}+\int_0^1 \int_0^1 \frac{y^2}{\sqrt{2-y^2}(2+y^2+y^2x^2)}dx\ dy\\ =&\ \frac{\pi^2}{36}+\int_0^1 \frac{\frac\pi4}{1+x^2}-\frac{\tan^{-1}\sqrt{2+x^2}}{(1+x^2)\sqrt{2+x^2}}\ dx\\ =&\ \frac{\pi^2}{36}+ \frac{\pi^2}{16}-\frac{5\pi^2}{96}= \frac{11\pi^2}{288}\\ \\ I_2=&\int_{0}^{1} \frac{1}{\left ( 1+y^2 \right )\sqrt{2+y^2} } \sin^{-1}\sqrt{\frac{1-y^2}{3} } \ \text{d}y \\ \overset{ibp}=&\int_0^1 \frac{y\tan^{-1}\frac y{\sqrt{2+y^2}}}{\sqrt{(1-y^2)(2+y^2)}}dy\\ =&\int_0^1 \int_0^1 \frac{y^2}{\sqrt{1-y^2}(2+y^2+y^2x^2)}dx\ dy\\ =&\ \frac\pi4 \int_0^1 \frac{1}{1+x^2}-\frac{\sqrt2}{(1+x^2)\sqrt{3+x^2}}\ dx\\ =&\ \frac{\pi^2}{8}-\frac{\pi}{2} \sin^{-1}\frac{1}{\sqrt{3}} \\ \\ I_3=& \int_{0}^{1} \frac{1}{( 1+y^2 )\sqrt{2+y^2} } \tan^{-1} \frac{\sqrt{(1-y^2 )\left ( 2+y^2 \right ) }} {2} \text{d}y\\ \overset{ibp}=&\int_0^1 \frac{2y(2y^2+1)\tan^{-1}\frac y{\sqrt{2+y^2}}}{(2-y^2)(3+y^2)\sqrt{(1-y^2)(2+y^2)}}dy\\ =&\int_0^1 \left(\frac{2y}{2-y^2}- \frac{2y}{3+y^2}\right)\frac{\tan^{-1}\frac y{\sqrt{2+y^2}}}{\sqrt{(1-y^2)(2+y^2)}}dy\\ =&\int_0^1\int_0^1 \left(\frac{1}{2-y^2}- \frac{1}{3+y^2}\right)\frac{2y^2}{(2+y^2+x^2y^2) \sqrt{1-y^2}}dx\ dy\\ =&\int_0^1\int_0^1 \frac{2}{(2+x^2) \sqrt{1-y^2}}\left(\frac{1}{2-y^2}- \frac{1}{2+y^2+x^2y^2}\right)dy\ dx\\ - &\int_0^1\int_0^1 \frac{2}{(1+3x^2) \sqrt{1-y^2}}\left(\frac{3}{3+y^2}- \frac{2}{2+y^2+x^2y^2}\right)dy\ dx\\ =& \ \frac\pi{\sqrt2}\int_0^1\bigg(1-\frac1{\sqrt{3+x^2}}\bigg)\frac1{2+x^2}dx \\ & -\frac\pi{2}\int_0^1\bigg(\sqrt3-\frac{2\sqrt2}{\sqrt{3+x^2}}\bigg)\frac1{1+3x^2}dx \\ =& \ \frac\pi2 \tan^{-1}\frac{1}{\sqrt{2} }-\frac\pi2 \tan^{-1}\frac{1}{2\sqrt{2} }-\frac{\pi^2}{6}+ \frac\pi2 \tan^{-1}{\sqrt{2} }\\ =&\ \pi\tan^{-1}\frac{1}{\sqrt{2} } -\frac{\pi^2}{6}\\ \end{aligned}