Definitely integrable

An alternative solution to Mr @Hrithik Thakur:

$\large \displaystyle I = \int_0^2 \frac{\ln (1 + 2x)}{1+x^2} dx$

Let $x = \tan (\theta)$ . Also $dx = \sec^2 (\theta) d \theta \rightarrow d \theta = \frac{dx}{1+x^2}$ and the limits of integration will be $0$ and $\arctan (2)$ . Let us call $\phi = \arctan (2)$ :

$\large \displaystyle I = \int_0^{\phi} \ln (1 + 2 \tan(\theta) ) d \theta$

$\large \displaystyle I = \int_0^{\phi} \ln \left ( \frac{\cos (\theta) + 2 \sin(\theta) )}{\cos(\theta)} \right ) d \theta$

$\large \displaystyle I = \int_0^{\phi} \ln (\cos (\theta) + 2 \sin(\theta)) d \theta - \int_0^{\phi} \ln (\cos (\theta)) d \theta$

Inside the $\ln$ of the first integral, we can multiply above and below by $\sqrt{5}$ :

$\large \displaystyle I = \int_0^{\phi} \ln \left [ \sqrt{5} \left ( \frac{\cos (\theta)}{\sqrt{5}} + 2 \frac{\sin(\theta)}{\sqrt{5}} \right ) \right ] d \theta - \int_0^{\phi} \ln (\cos (\theta)) d \theta$

Notice that if $\phi = \arctan (2)$ , $\cos(\phi) = \frac{1}{\sqrt{5}}$ and $\sin(\phi) = \frac{2}{\sqrt{5}}$ :

$\large \displaystyle I = \int_0^{\phi} \ln \left [ \sqrt{5} \left (\cos (\theta) \cos(\phi) + \sin(\theta)\sin(\phi) \right ) \right ] d \theta - \int_0^{\phi} \ln (\cos (\theta)) d \theta$

$\large \displaystyle I = \int_0^{\phi} \ln(\sqrt{5})d \theta + \int_0^{\phi} \ln(\cos(\theta - \phi))d \theta - \int_0^{\phi} \ln (\cos (\theta)) d \theta$

In the second integral make $u = \phi - \theta$ . Then $du = - d \theta$ :

$\large \displaystyle I = \int_0^{\phi} \ln(\sqrt{5})d \theta + \int_{\phi} ^0- \ln(\cos(-u))du - \int_0^{\phi} \ln (\cos (\theta)) d \theta$

Since cosine is even:

$\large \displaystyle I = \int_0^{\phi} \ln(\sqrt{5})d \theta + \int_0^{\phi} \ln(\cos(u))du - \int_0^{\phi} \ln (\cos (\theta)) d \theta$

The second and the third integral cancel out and:

$\large \displaystyle I = \int_0^{\phi} \ln(\sqrt{5})d \theta$

$\color{#20A900} \boxed{ \large \displaystyle I = \ln(\sqrt{5}) \arctan(2) }$

Thus:

$\color{#3D99F6} \large \displaystyle a = 2, b = 5, \boxed{ \large \displaystyle a^4 + b^4 = 641}$

Elaboration on @Hrithik Thakur's solution.

$\begin{aligned} I(p) & = \int_0^2 \frac {\ln (1+px)}{1+x^2} dx \\ \frac {\partial I(p)}{\partial p} & = \int_0^2 \frac x{(1+px)(1+x^2)} dx \\ & = \frac 1{p^2+1}\int_0^2 \left(\frac {x+p}{x^2+1} - \frac p{px+1} \right) dx \\ & = \frac 1{p^2+1} \left[\frac {\ln (x^2+1)}2 + p\tan^{-1} x - \ln (px+1) \right]_0^2 \\ & = \frac {\ln \sqrt 5 + p\tan^{-1}2 - \ln (2p+1)}{p^2+1} \\ \implies I(p) & = \int \frac {\ln \sqrt 5 + p\tan^{-1}2 - \ln (2p+1)}{p^2+1} dp \\ & = \ln \sqrt 5 \tan^{-1} p + \ln \sqrt{p^2+1} \tan^{-1}2 - \int \frac {\ln (2p+1)}{p^2+1} dp + C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ & = \ln \sqrt 5 \tan^{-1} p + \ln \sqrt{p^2+1} \tan^{-1}2 - \int \frac {\ln (2p+1)}{p^2+1} dp & \small \color{#3D99F6} \text{Since } I(0) = 0 \implies C = 0 \\ I(2) & = \ln \sqrt 5 \tan^{-1} 2 + \ln \sqrt 5 \tan^{-1}2 - I(2) \\ & = \tan^{-1} 2 \ln \sqrt 5 \end{aligned}$

Therefore $a^4 + b^4 = 2^4+5^4 = \boxed{641}$ .

Hint: generalize the given integrand as $\int_{0}^{2} \frac{\ln(1+px)}{1+x^{2}}dx$ then use differentiation under integration.

Here's a more general form :-

$\int_{0}^{p} \frac{ln(1+px)}{1+x^{2}} dx = I$

Let $x=tan(t)$

so the integral transforms to:-

$\int_{0}^{arctan(p)}ln(1+ptan(t))dt = \int_{0}^{arctan(p)}ln(1+ptan(arctan(p) - t))dt$

$= \int_{0}^{arctan(p)}{ln(1+p\frac{(p-tan(t))}{1+ptan(t)})}dt$

$= \int_{0}^{arctan(p)}{ln(\frac{1+p^{2}}{1+ptan(t)})}dt$

So $2I = \int_{0}^{arctan(p)}ln(1+p^{2}) dt$

$2I = arctan(p)ln(1+p^{2})$

$I = arctan(p)ln(\sqrt{1+p^{2}})$

Plug in $p=2$ to get the answer .