I can! Can you? (2)

$\begin{aligned} I & = \int_0^\infty \frac 1{2x^4+x^2+1} dx \\ & = \int_0^\infty \frac 1{2\left(x^2 + \frac 14 - \frac {\sqrt 7}4i\right)\left(x^2 + \frac 14 + \frac {\sqrt 7}4i\right)} \\ & = \frac i{\sqrt 7} \int_0^\infty \left(\frac 1{x^2 + \frac 14 + \frac {\sqrt 7}4i} - \frac 1{x^2 + \frac 14 - \frac {\sqrt 7}4i}\right) dx \\ & = \frac {2i}{\sqrt{7+7\sqrt 7i}} \int_0^\infty \frac 1{\left(\frac {2x}{\sqrt{1+\sqrt 7i}}\right)^2 + 1} dx - \frac {2i}{\sqrt{7-7\sqrt 7i}} \int_0^\infty \frac 1{\left(\frac {2x}{\sqrt{1-\sqrt 7i}}\right)^2 + 1} dx \\ & = \frac {\pi i}{\sqrt 7} \left(\frac 1{\sqrt{1+\sqrt 7i}} - \frac 1{\sqrt{1-\sqrt 7i}} \right) \\ & = \frac {\pi i}{\sqrt{56}} \left(\sqrt{1-\sqrt 7i} - \sqrt{1+\sqrt 7i} \right) & \small \color{#3D99F6} \text{Using Euler's formula: } e^{i\theta} = \cos \theta + i \sin \theta \\ & = \frac {\pi i}{\sqrt{14\sqrt 2}} \left(e^{-\frac 12 \tan^{-1} \sqrt 7} - e^{\frac 12 \tan^{-1} \sqrt 7}\right) \\ & = \frac {2 \pi}{\sqrt{14\sqrt 2}} \sin \left(\frac {\tan^{-1} \sqrt 7}2\right) \\ & = \pi \left(\frac {2\sqrt 2-1}{\sqrt{14\sqrt 2}\sqrt{4-\sqrt 2}} \right) \\ & = \pi \sqrt{\frac 1{7\sqrt 2}-\frac 1{28}} \end{aligned}$

Therefore, $a+b+c = 7+2+28 = \boxed{37}$ .

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Reference: Euler's formula

alright!!for those who think indefinte integral seems to be hard make substitution

x = t/ ( (2)^(0.25) ) ..Now i believe most of u can do the indefinte integral .the u will have (integral is there on RHS)

2^0.25 I = 1 / ( t^4+ t^2/sqrt2 + 1 )

2^0.25 I =(1/t^2) / ( t^2 + t^(-2) + 2^(-0.5) )

2^(1.25 )I = ( (1+1/t^2) - (1-1/t^2) )/ ( t^2 + t^(-2) + 2^(-0.5) )

Now just split it and make complete the square in denominator