Inspired by Karthik Sharma

$\begin{array} {rll} I & = \displaystyle \int_0^\infty \frac{x^4}{(1+x^4)^2} dx & \small \color{#3D99F6}{\text{Let } x = \tan^{\frac{1}{2}} \theta \quad \Rightarrow dx = \frac{1}{2} \tan^{-\frac{1}{2}} \theta \sec^2 \theta \space d\theta} \\ & = \displaystyle \int_0^{\frac{\pi}{2}} \frac{\tan^2 \theta\tan ^{-\frac{1}{2}} \theta \sec^2 \theta }{2(\sec^2 \theta)^2} d \theta \\ & = \displaystyle \frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{\tan^{\frac{3}{2}} \theta}{\sec^2 \theta} d \theta \\ & = \displaystyle \frac{1}{2} \int_0^{\frac{\pi}{2}} \sin^{\frac{3}{2}} \theta \cos^{\frac{1}{2}} \theta d \theta & \small \color{#3D99F6} { \displaystyle \int_0^{\frac{\pi}{2}} \sin^{x} \theta \cos^{y} \theta d \theta = \frac{1}{2} B \left(\frac{1}{2}(x+1), \frac{1}{2}(y+1) \right)} \\ & = \frac{1}{4} B \left(\frac{5}{4}, \frac{3}{4}\right) & \small \color{#3D99F6}{B(x,y) = \dfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}} \\ & = \dfrac{\color{#3D99F6}{\Gamma \left(\frac{5}{4} \right)} \Gamma \left( \frac{3}{4} \right)} {4 \Gamma \left( \frac{5}{4} + \frac{3}{4} \right)} & \small \color{#3D99F6}{\Gamma (1+x) = x \Gamma(x)} \\ & = \dfrac{\color{#3D99F6}{\frac{1}{4} \Gamma \left(\frac{1}{4} \right)} \Gamma \left( \frac{3}{4} \right)} {4 \color{#D61F06} {\Gamma \left( 2 \right)}} & \small \color{#D61F06}{\Gamma(n) = (n-1)!} \\ & = \dfrac{\color{#3D99F6}{\Gamma \left(\frac{1}{4} \right) \Gamma \left( \frac{3}{4} \right)}} {16 \color{#D61F06} {(1)}} & \small \color{#3D99F6}{\Gamma(x) \Gamma(1-x) = \dfrac{\pi}{\sin (\pi x)}, \quad x = \frac{3}{4}} \\ & = \dfrac{\sqrt{2}\pi}{16} = \dfrac{\pi}{8\sqrt{2}} \end{array}$

$\Rightarrow 10a + b = \boxed{82}$

Take $1+x^4=z$ $$ Differentiating we get, $\dfrac{dz}{dx}=4x^3$ .Now,the integral becomes, $I=\int_{0}^{\infty}\dfrac{x^4}{(1+x^4)^{2}}\\ =\dfrac{1}{4}\int_{1}^{\infty}\dfrac{(z-1)^{\frac{1}{4}}}{z^2}dz$ .Now,taking $z=\sec^2\theta$ , $$ differentiating we get, $\dfrac{dz}{d\theta}=2\sec^2\theta\tan\theta$ ,substituting we get, $\dfrac{1}{2}\int_{0}^{\frac{\pi}{2}}\dfrac{\sqrt{\tan\theta}\sec^2\theta\tan\theta}{\sec^4\theta}d\theta\\ =\dfrac{1}{2}\int_{0}^{\frac{\pi}{2}}\dfrac{\tan^{\frac{3}{2}}\theta}{\sec^2\theta}$ .Converting it into $\sin$ and $cos$ ,we get, $I=\dfrac{1}{2}\int_{0}^{\frac{\pi}{2}}\sin^{\frac{3}{2}}\theta\cos^{\frac{1}{2}}\theta d\theta\\ I=\dfrac{1}{2}\int_{0}^{\frac{\pi}{2}}\cos^{\frac{3}{2}}\theta\sin^{\frac{1}{2}}\theta d\theta$ .Adding we get that, $2I=\dfrac{1}{2}\int_{0}^{\frac{\pi}{2}}\sqrt{\sin\theta\cos\theta}(\sin\theta+\cos\theta)d\theta$ .Making the substitution $\sin\theta-\cos\theta=y\Longrightarrow \dfrac{1-y^2}{2}=\sin\theta\cos\theta$ ,differentiating we get, $$ $\dfrac{dy}{d\theta}=\cos\theta+\sin\theta$ .Substituting we get, $2I=\dfrac{1}{2}\int_{-1}^{1}\sqrt{1-y^2}dy\\ =\int_{0}^{1}\sqrt{1-y^2}dy$ .Substituting $y=\sin\phi$ ,we get $$ $\dfrac{d\phi}{dy}=\cos\phi$ .Substituting in the integral we get, $2I=\int_{0}^{\frac{\pi}{2}}\cos^2\phi d\phi$ .This is a simple one I leave it to you.The final answer that you should get is $\color{#D61F06}{\boxed{\large{I=\displaystyle\int_{0}^{\infty}\dfrac{x^4}{(1+x^4)^{2}}=\dfrac{\pi}{8\sqrt{2}}}}}$ .And done!

Alternate Solution:

$I=\int _{ 0 }^{ \infty }{ \frac { { x }^{ 4 } }{ { \left( 1+{ x }^{ 4 } \right) }^{ 2 } } dx } \\ =\frac { 1 }{ 4 } \int _{ 0 }^{ \infty }{ \frac { { x }^{ \frac { 1 }{ 4 } } }{ { \left( 1+{ x } \right) }^{ 2 } } dx } \\ =\frac { 1 }{ 4 } M\left\{ \frac { 1 }{ { \left( 1+{ x } \right) }^{ 2 } } \right\} \left( \frac { 1 }{ 4 } \right)$

where M{f(x)}(s) is Mellin Transform of f(x) over s.

Therefore, by Ramanujan's Master Theorem,

$I=\frac { \Gamma \left( \frac { 1 }{ 4 } \right) \Gamma \left( \frac { 3 }{ 4 } \right) }{ 16 }$

$I\left( a \right) =\int _{ 0 }^{ \infty } \frac { -1 }{ (1+ax^{ 4 }) } dx\\ let\quad x=u\cdot a^{ \frac { -1 }{ 4 } }\Rightarrow dx=du\cdot a^{ \frac { -1 }{ 4 } }\\ I\left( a \right) =-a^{ \frac { -1 }{ 4 } }\int _{ 0 }^{ \infty } \frac { 1 }{ 1+u^{ 4 } } du=-a^{ \frac { -1 }{ 4 } }\frac { \pi }{ 2\sqrt { 2 } } \\ \frac { d }{ da } I\left( a \right) =\int _{ 0 }^{ \infty } \frac { x^{ 4 } }{ (1+ax^{ 4 })^{ 2 } } dx=\frac { \pi }{ 2\sqrt { 2 } } \cdot \frac { 1 }{ 4 } a^{ \frac { -5 }{ 4 } }\\ \frac { d }{ da } I\left( 1 \right) =\int _{ 0 }^{ \infty } \frac { x^{ 4 } }{ (1+x^{ 4 })^{ 2 } } dx=\frac { \pi }{ 8\sqrt { 2 } }$

Integrating by parts: $\displaystyle I = \int_0^\infty\frac{x^4}{(1+x^4)^2}dx = \frac{-x}{4(1+x^4)}\bigg|^\infty_0+\frac1{4}\int_0^\infty\frac1{1+x^4}dx$

$\displaystyle\int_{0}^{\infty}\frac{1}{x^n+1}dx = \int_{0}^{\infty}\int_{0}^{\infty}e^{-t(1+x^n)}dxdt$

Sub $\displaystyle u = x^nt\quad du = nx^{n-1}t\,dx$

$\displaystyle\frac{1}{n}\int_{0}^{\infty}e^{-t}t^{\frac{-1}{n}}dt\int_{0}^{\infty}u^{\frac{1}{n}-1}e^{-u}du=\frac{\Gamma(1-\frac{1}{n})\Gamma(\frac{1}{n})}{n} = \frac{\pi}{n\sin{\frac{\pi}{n}}}$

$\displaystyle I = \frac1{4}\int_0^\infty\frac1{1+x^4}dx = \boxed{\frac{\pi}{8\sqrt{2}}}$