<s>[INST] How to prove $\int_0^1\frac{x^3\arctan x}{(3-x^2)^2}\frac{\mathrm dx}{\sqrt{1-x^2}}=\frac{\pi\sqrt{2}}{192}\left(18-\pi-6\sqrt{3}\,\right)$? How to prove the following result?

$$\int_0^1\frac{x^3\arctan x}{(3-x^2)^2}\frac{\mathrm dx}{\sqrt{1-x^2}}=\frac{\pi\sqrt{2}}{192}\left(18-\pi-6\sqrt{3}\,\right)$$

For my part no idea?
 [/INST] $$\newcommand{\b}[1]{\left(#1\right)}
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This belongs to a large category of functions that are easily solvable with integration by parts like:

$$\int\frac{x\arctan(x)}{\sqrt{1+x^2}}\d x=\sqrt{1+x^2}\arctan(x)-\ln|x+\sqrt{x^2+1}|+c$$


So let our functions be:
$$f(x)=\f{x^3}{(3-x^2)^2\s{1-x^2}};g(x)=\arctan(x)\\\int f(x)g(x)\d x=g(x)\int f(x)\d x-\int g'(x)\u{\int f(x)\d x}_{h(x)}\d x$$
And(Beware I used Reduction Formula for evaluating last integral!):
$$\int f(x)\d x=\f12\int\f{t\d t}{(3-t)^2\s{1-t}}\quad(t=x^2)\\
=\int\f{u^2-1}{(u^2+2)^2}\d u=\int\f1{(u^2+2)}\d u-3\int\f{1}{(u^2+2)^2}\d u\quad(u^2=1-t)\\
=\frac1{\sqrt2}\arctan\frac u{\sqrt2}-3\b{\frac u{4(u^2+2)}+\f1{4\s2}\arctan\f u{\s2}}+{\cal C}\\
=\f1{4\s2}\arctan\s{\f{1-x^2}2}-\f34\f{\s{1-x^2}}{3-x^2}+{\cal C}=\t{(let) }h(x)$$
And thus:
$$\int g'(x)h(x)\d x=\f1{4\s2}\int \f1{1+x^2}\arctan\s{\f{1-x^2}2}\d x-\f34\int \f{\s{1-x^2}}{(3-x^2)(1+x^2)}\d x$$
So:
$$I=\int_0^1f(x)g(x)\d x=\b{\arctan (x)\b{\f1{4\s2}\arctan\s{\f{1-x^2}2}-\f34\f{\s{1-x^2}}{3-x^2}}}_0^1\\-\f1{4\s2}\int_0^1 \f1{1+x^2}\arctan\s{\f{1-x^2}2}\d x+\f34\int_0^1 \f{\s{1-x^2}}{(3-x^2)(1+x^2)}\d x\\
I=\f34\u{\int_0^1 \f{\s{1-x^2}}{(3-x^2)(1+x^2)}\d x}_{I_1}-\f1{4\s2}\int_0^1 \f1{1+x^2}\arctan\s{\f{1-x^2}2}\d x$$
Now:
$$I_1=\int_0^\infty\f{\d y}{4y^4+8y^2+3}\quad\b{x=\f{y^2}{y^2+1}\t{ or }y=\f{x}{\s{1-x^2}}=\tan\arcsin x \\\t{ which can also be obtained after two sunstitutions viz., }x=\sin z\t{, then }y=\tan z}\\
\f12\b{\f1{2y^2+1}-\f1{2y^2+3}}_0^{\infty}\\
=\f{3-\s3}{24}\pi\s2=\f{3-\s3}{24}\pi\s2$$
Still working on the last integral!!
BTW the final form is:
$$I=\frac{\pi\sqrt{2}}{192}(18-6\sqrt{3})-\f1{4\s2}\int_0^1 \f1{1+x^2}\arctan\s{\f{1-x^2}2}\d x$$
So it remains to prove:
$$\int_0^1 \f1{1+x^2}\arctan\s{\f{1-x^2}2}\d x=\pi^2/24$$
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