Happy Birthday Rajdeep!

$\large \mathfrak{I}= \int_{-1/\sqrt{3}}^{1/\sqrt{3}} \dfrac{x^2(1-x^2)}{(e^x+1)(1+x^2)^4} \ dx \$ $\mathfrak{I}= \int_{-1/\sqrt{3}}^{1/\sqrt{3}} e^x(\dfrac{x^2(1-x^2)}{(e^x+1)(1+x^2)^4} )\ dx \$ $\small{(\because\color{#D61F06}{\int_{a}^{b}f(x)\, dx=\int_{a}^{b}f(a+b-x)\, dx})}$ $2\mathfrak{I}= \int_{-1/\sqrt{3}}^{1/\sqrt{3}} \dfrac{x^2(1-x^2)}{(1+x^2)^4} \ dx \$ $\Rightarrow \mathfrak{I}=\int_{0}^{1/\sqrt{3}} \dfrac{x^2(1-x^2)}{(1+x^2)^4} \ dx \$ Substitute $x=tany$ and use $\tan y=\frac{\sin y}{\cos y}$ and $\sec y=\frac{1}{\cos y}$ followed by $2 \sin y\cos y=\sin 2y$ and $\cos^2 y-\sin^2 y=\cos 2y$ . $\large \mathfrak{I}=\dfrac{1}{4} \int_{0}^{\frac{\pi}{6}}(\sin^2 2y ) (\cos 2y) \,dy$ $\large =\dfrac{1}{8}(\dfrac{(\sin 2 y)^3}{3})|_{\small{0}}^{\small{\frac{\pi}{6}}}$ $\Large =\dfrac{\sqrt 3}{64}$ $\huge \therefore ~3+64=\boxed{\color{#007fff}{67}}$

$\large \mathfrak{I}= \int_{-1/\sqrt{3}}^{1/\sqrt{3}} \dfrac{x^2(1-x^2)}{(e^x+1)(1+x^2)^4} \ dx \$ Same method as shown by Rishabh above this can be simplified to: $\Rightarrow \mathfrak{I}=\int_{0}^{1/\sqrt{3}} \dfrac{x^2(1-x^2)}{(1+x^2)^4} \ dx \$

$\Rightarrow \mathfrak{I}=\int_{0}^{1/\sqrt{3}} \dfrac{(\frac{1}{x^2}-1)}{(\frac{1}{x}+x)^4} \ dx \$

Let $x+\frac{1}{x} =t$

$(1-\frac{1}{x^2})dx =dt$ $\Rightarrow \mathfrak{I}=\int_{0}^{1/\sqrt{3}}\frac{-1}{t^4}dt$ So $\Rightarrow \mathfrak{I}=\frac{1}{3(x+\frac{1}{x})^3}$ putting limits we can get the answer

First of all, Happy Birthday @Rajdeep Dhingra :)

Now, let $I=\int _{ \frac { -1 }{ \sqrt { 3 } } }^{ \frac { 1 }{ \sqrt { 3 } } }{ \frac { { x }^{ 2 }(1-{ x }^{ 2 }) }{ ({ e }^{ x }+1){ (1+{ x }^{ 2 }) }^{ 4 } } } dx$

Using the property: $\int _{ a }^{ b }{ f(x) } dx\quad =\quad \int _{ a }^{ b }{ f(a+b-x) } dx$ , we get

$I=\int _{ \frac { -1 }{ \sqrt { 3 } } }^{ \frac { 1 }{ \sqrt { 3 } } }{ \frac { { x }^{ 2 }(1-{ x }^{ 2 }) }{ ({ e }^{ -x }+1){ (1+{ x }^{ 2 }) }^{ 4 } } } dx\quad =\int _{ \frac { -1 }{ \sqrt { 3 } } }^{ \frac { 1 }{ \sqrt { 3 } } }{ \frac { { x }^{ 2 }(1-{ x }^{ 2 }){ e }^{ x } }{ ({ e }^{ x }+1){ (1+{ x }^{ 2 }) }^{ 4 } } } dx$

Adding both the values for $I$ , we get:

$2I\quad =\int _{ \frac { -1 }{ \sqrt { 3 } } }^{ \frac { 1 }{ \sqrt { 3 } } }{ \frac { { x }^{ 2 }(1-{ x }^{ 2 }) }{ { (1+{ x }^{ 2 }) }^{ 4 } } } dx$

Now, substituting $x=\tan { \theta }$ and $dx=\sec ^{ 2 }{ \theta } d\theta$ , we can write $2I$ as:

$2I\quad =\int _{ \frac { -\pi }{ 6 } }^{ \frac { \pi }{ 6 } }{ \frac { \tan ^{ 2 }{ \theta } (1-\tan ^{ 2 }{ \theta } ) }{ { (1+\tan ^{ 2 }{ \theta } ) }^{ 4 } } } \sec ^{ 2 }{ \theta } d\theta$

Using the identity: $1+\tan ^{ 2 }{ \theta } =\sec ^{ 2 }{ \theta }$ and $1-\tan ^{ 2 }{ \theta } =\frac { \cos { 2\theta } }{ \cos ^{ 2 }{ \theta } }$ , we get:

$2I\quad =\int _{ \frac { -\pi }{ 6 } }^{ \frac { \pi }{ 6 } }{ \frac { \tan ^{ 2 }{ \theta } \frac { \cos { 2\theta } }{ \cos ^{ 2 }{ \theta } } }{ { (1+\tan ^{ 2 }{ \theta } ) }^{ 3 } } } d\theta \\ \Rightarrow \quad 2I\quad =\int _{ \frac { -\pi }{ 6 } }^{ \frac { \pi }{ 6 } }{ \frac { \frac { \sin ^{ 2 }{ \theta } }{ \cos ^{ 2 }{ \theta } } \frac { \cos { 2\theta } }{ \cos ^{ 2 }{ \theta } } }{ ({ \sec ^{ 2 }{ \theta } ) }^{ 3 } } } d\theta$

Further simplifying, we get:

$2I\quad =\int _{ \frac { -\pi }{ 6 } }^{ \frac { \pi }{ 6 } }{ \frac { \sin ^{ 2 }{ \theta } }{ \cos ^{ 2 }{ \theta } } \frac { \cos { 2\theta } }{ \cos ^{ 2 }{ \theta } } \cos ^{ 6 }{ \theta } } d\theta \\ \Rightarrow \quad 2I\quad =\int _{ \frac { -\pi }{ 6 } }^{ \frac { \pi }{ 6 } }{ (\sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta } )\cos { 2\theta } } d\theta \\ \Rightarrow \quad 2I\quad =\frac { 1 }{ 4 } \int _{ \frac { -\pi }{ 6 } }^{ \frac { \pi }{ 6 } }{ \sin ^{ 2 }{ 2\theta } \cos { 2\theta } } d\theta \\ \Rightarrow \quad 2I\quad =\frac { 1 }{ 4 } \frac { 1 }{ 2 } { \frac { \sin ^{ 3 }{ 2\theta } }{ 3 } }]_{ \frac { -\pi }{ 6 } }^{ \frac { \pi }{ 6 } }\\ \Rightarrow \quad I\quad =\frac { 1 }{ 48 } (\sin ^{ 3 }{ \frac { \pi }{ 3 } } -\sin ^{ 3 }{ \frac { -\pi }{ 3 } ) } \\ \Rightarrow \quad I\quad =\frac { 2 }{ 48 } \frac { 3\sqrt { 3 } }{ 8 } =\frac { \sqrt { 3 } }{ 64 }$

Now, comparing, we get $a=3$ , and $b=64$ . So, the answer, $a+b=67$ . :)

We use Weierstrass substitution where $x=\tan\dfrac{y}{2} \Rightarrow y=2\arctan(x)$ and then $\sin x=\dfrac{2x}{1+x^2}$ , $\cos x=\dfrac{1-x^2}{1+x^2}$ , $dy=\dfrac{2}{1+x^2} dx$ . Now note that $\displaystyle\int_{-1/\sqrt{3}}^{1/\sqrt{3}} \dfrac{x^2(1-x^2)}{(e^x+1)(1+x^2)^4} \, dx$ can be written as:

$\dfrac{1}{8}\int_{-1/\sqrt{3}}^{1/\sqrt{3}} \dfrac{1}{1+e^x} \times \dfrac{4x^2}{(1+x^2)^2} \times \dfrac{1-x^2}{1+x^2} \times \dfrac{2}{1+x^2} dx$

Now in this integral , as $x=1/\sqrt{3} \ , \ y=\pi/3$ and $x=-1/\sqrt{3} \ , \ y=-\pi/3$ and we change the limits accordingly.Using Weierstrass substitution we transform it to:

$\dfrac{1}{8}\int_{-\pi/3}^{\pi/3} \dfrac{\sin^2 y\cos y}{1+e^y} \ dy$

Now since $\dfrac{\sin^2 y\cos y}{1+e^y}$ is an even function we use the property $\displaystyle\int\limits_{-a}^a\frac{E(x)}{e^x+1}\,\mathrm dx=\int_0^a E(x)\,\mathrm dx$ and then we have:

$\dfrac{1}{8}\int_{0}^{\pi/3} \sin^2 y\cos y \ dy$

Now using the substitution $u=\sin y$ , $du=\cos y dy$ and required changes in limits , we have:

$\dfrac{1}{8}\int_{0}^{\sqrt{3}/2} u^2 \ du$

$=\dfrac{1}{8} \times \dfrac{u^3}{3} \bigg|_{0}^{\sqrt{3}/2}=\dfrac{3^{3/2}}{64\times 3}=\boxed{\dfrac{\sqrt{3}}{64}}$

Hence , $a=3 \ , \ b=64 \ , \ a+b=3+64=\boxed{67}$