A calculus problem by Isay Katsman

$\large \displaystyle I = \int_0^{\sqrt{2}} \frac{1}{x^4 + 16} dx$

By partial fractions:

$\large \displaystyle I = \frac{1}{16\sqrt{2}} \int_0^{\sqrt{2}} \left [ \frac{x + 2\sqrt{2}}{x^2 + 2\sqrt{2}x + 4} - \frac{x - 2\sqrt{2}}{x^2 - 2\sqrt{2}x + 4} \right ] dx$

$\large \displaystyle I = \frac{1}{16\sqrt{2}} \int_0^{\sqrt{2}} \left [ \frac{x + 2\sqrt{2}}{ ( x + \sqrt{2})^2 + 2} - \frac{x - 2\sqrt{2}}{( x - \sqrt{2})^2 + 2} \right ] dx$

Multuplying below and above by $2$ :

$\large \displaystyle I = \frac{1}{32\sqrt{2}} \int_0^{\sqrt{2}} \left [ \frac{2x + 4\sqrt{2}}{ ( x + \sqrt{2})^2 + 2} - \frac{2x - 4\sqrt{2}}{( x - \sqrt{2})^2 + 2} \right ] dx$

$\large \displaystyle I = \frac{1}{32\sqrt{2}} \left [ \int_0^{\sqrt{2}} \frac{2x + 2\sqrt{2}}{ ( x + \sqrt{2})^2 + 2} dx + \int_0^{\sqrt{2}} \frac{2\sqrt{2}}{ ( x + \sqrt{2})^2 + 2} dx - \int_0^{\sqrt{2}} \frac{2x - 2\sqrt{2}}{ ( x - \sqrt{2})^2 + 2} dx + \int_0^{\sqrt{2}} \frac{2\sqrt{2}}{ ( x - \sqrt{2})^2 + 2} dx \right ]$

$\large \displaystyle I = \frac{1}{32\sqrt{2}} \left [ \ln[ (x+\sqrt{2})^2 + 2] \Bigg |_0^{\sqrt{2}} + \int_0^{\sqrt{2}} \frac{\sqrt{2}}{ \left ( \frac{x}{\sqrt{2}} + 1 \right )^2 + 1} dx - \ln[ (x-\sqrt{2})^2 + 2] \Bigg |_0^{\sqrt{2}} + \int_0^{\sqrt{2}} \frac{\sqrt{2}}{ \left ( \frac{x}{\sqrt{2}} - 1 \right )^2 + 1} dx \right ]$

$\large \displaystyle I = \frac{1}{32\sqrt{2}} \left [ (\ln(10) - \ln(4) ) + \sqrt{2} \cdot \sqrt{2} \tan^{-1} \left ( \frac{x}{\sqrt{2}} + 1 \right ) \Bigg |_0^{\sqrt{2}} - (\ln(2) - \ln(4) ) + \sqrt{2} \cdot \sqrt{2} \tan^{-1} \left ( \frac{x}{\sqrt{2}} - 1 \right ) \Bigg |_0^{\sqrt{2}} \right ]$

$\large \displaystyle I = \frac{1}{32\sqrt{2}} \left [ \ln(10) - \ln(2) + 2(\tan^{-1} (2) - \tan^{-1} (1) ) + 2(\tan^{-1} (0) - \tan^{-1} (-1) ) \right ]$

$\color{#20A900} \boxed{ \large \displaystyle I = \frac{\ln(5) + 2\tan^{-1}(2)}{32\sqrt{2}} }$

Then:

$\color{#3D99F6} \large \displaystyle a = 5, b = 2, c = 2, d = 32, e = 2, \boxed{\large \displaystyle abcde = 1280}$