How would you solve it -Part 2

Similar solution to Kartik Sharma 's, given more details.

$\begin{aligned} I & = \int_0^{\infty} \left(\sqrt{1+x^4} - x^2 \right) dx & \small \color{#3D99F6}{\left[\text{Let } x^2 = \tan \theta \quad \Rightarrow dx = \frac{\sec^2 \theta}{2\sqrt{\tan \theta}} d \theta\right]} \\ & = \small \int_0^{\frac{\pi}{2}} \left(\sqrt{1+\tan^2 \theta} - \tan \theta \right)\dot{} \frac{\sec^2 \theta}{2\sqrt{\tan \theta}} d \theta \\ & = \small \frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{\sec^3 \theta - \tan \theta \sec^2 \theta}{\sqrt{\tan \theta}} d \theta \\ & = \small \frac{1}{2} \int_0^{\frac{\pi}{2}} \color{#3D99F6} {\sin^{-\frac{1}{2}} \theta \cos^{-\frac{5}{2}} \theta} - \color{#3D99F6} {\sin^{\frac{1}{2}} \theta \cos^{-\frac{5}{2}} \theta} \space d \theta & \small \color{#3D99F6}{\left[ \int_0^{\frac{\pi}{2}} sin^{m}\theta \cos^{n} d\theta = \frac{1}{2}B \left( \frac{1}{2}(m+1), \frac{1}{2}(n+1) \right)\right]} \\ & = \small \dfrac{1}{4} \left[ \color{#3D99F6} {B\left(\frac{1}{4} \right) B\left(-\frac{3}{4} \right)} - \color{#D61F06}{B\left(-\frac{1}{4} \right) B\left(-\frac{3}{4} \right)} \right] & \small \left[ \color{#3D99F6}{B(x,y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}} \quad \color{#D61F06}{B\left(-\frac{1}{4} \right) B\left(-\frac{3}{4} \right) = 0} \right] \\ & = \small \frac{1}{4} \left[ \frac{\Gamma \left( \frac{1}{4} \right) \color{#3D99F6}{\Gamma \left(- \frac{3}{4} \right)}} {\color{#D61F06}{\Gamma \left( \frac{1}{4} - \frac{3}{4} \right)}} - 0 \right] & \small \color{#3D99F6} {\left[ \Gamma(1+x)= x \Gamma(x), \quad x = - \frac{3}{4} \quad \Rightarrow \Gamma \left(-\frac{3}{4} \right) = - \frac{4}{3} \Gamma \left(\frac{1}{4} \right)\right]} \\ & = \small \frac{1}{4} \left[ \frac{\Gamma \left( \frac{1}{4} \right) \color{#3D99F6}{\left( - \frac{4}{3} \Gamma \left(\frac{1}{4} \right) \right)}} {\color{#D61F06}{\Gamma \left( - \frac{1}{2} \right)}} \right] & \small \color{#D61F06} {\left[ \Gamma(1+x)= x \Gamma(x), \space x = - \frac{1}{2} \space \Rightarrow \Gamma \left(-\frac{1}{2} \right) = - 2 \Gamma \left(\frac{1}{2} \right) = - 2\sqrt{\pi}\right]} \\ & = \dfrac{\left(\Gamma \left( \frac{1}{4} \right) \right)^2}{6\sqrt{\pi}} \end{aligned}$

Therefore, $(a+b+c+2d)^2 = \left(2+4+6+2\times \frac{1}{2} \right)^2 = \boxed{169}$

Nice problem!

$\displaystyle \int_{0}^{\infty}{\left(\sqrt{1+x^4} - x^2\right) \ dx}$

Substitute $x^2 = \tan\theta$ , $\displaystyle dx = \frac{\sec^2\theta \ d\theta}{2\sqrt{\tan\theta}}$

$\displaystyle = \int_{0}^{\pi/2}{\frac{\sec\theta - \tan\theta}{2\sqrt{\tan\theta}} \sec^2\theta \ d\theta}$

$\displaystyle = \frac{1}{4} 2\int_{0}^{\pi/2}{\left({\cos}^{-5/2}\theta {\sin}^{-1/2}\theta - {\cos}^{-5/2}\theta {\sin}^{1/2}\theta\right) \ d\theta}$

$\displaystyle = \frac{1}{4} 2\int_{0}^{\pi/2}{\left({\cos}^{2(-3/4) -1}\theta {\sin}^{2(1/4)-1}\theta - {\cos}^{2(-3/4)-1}\theta {\sin}^{2(3/4)-1}\theta\right) \ d\theta}$

and we know that $\displaystyle 2 \int_{0}^{\pi/2}{\left({\cos}^{2x -1}\theta {\sin}^{2y-1}\theta\right) \ d\theta} = B(x,y)$ , $\displaystyle B(x,y)$ is the Beta function.

and hence

$\displaystyle = \frac{1}{4} \left(B\left(\frac{-3}{4}, \frac{1}{4}\right) - B\left(\frac{-3}{4}, \frac{3}{4}\right)\right) = - \frac{\Gamma\left(\frac{-3}{4}\right) \Gamma\left(\frac{1}{4}\right)}{8\sqrt{\pi}} = \frac{4}{3} \frac{\Gamma\left(\frac{1}{4}\right)^2}{8\sqrt{\pi}} = \frac{\Gamma\left(\frac{1}{4}\right)^2}{6\sqrt{\pi}}$