Your equation has a very simple solution given by
$$i(t)=Ae^{-\frac{R}{L}t}+\frac{E}{R}$$
being $A$ an arbitrary constant. Now, take a look at your Laplace transform
$$I(s)=\frac{E}{sL\left(s+\frac{R}{L}\right)}$$
and you will recognize two poles for $s=0$ and $s=-\frac{R}{L}$. The first pole will give rise to the constant term while the other one is just the exponential contribution. Indeed, here you can find a table of well-known transformed functions. This is essential in circuit analysis. Your case is the "exponential approach".
The integral wouldn't be so bad. The identity gives
$$\sin(2t)\int_0^t \frac{\cos(2u)}{4+\cos(2u)}\:du + \cos(2t)\int_0^t \frac{-\sin(2u)}{4+\cos(2u)}\:du$$
The second integral is easy enough
$$\cos(2t)\int_0^t \frac{-\sin(2u)}{4+\cos(2u)}\:du = \frac{1}{2}\cos(2t)\log(4+\cos(2t))$$
The trick to the second integral is to split it up
$$\sin(2t)\int_0^t \frac{4+\cos(2u)-4}{4+\cos(2u)}\:du = t\sin(2t) - 4\sin(2t)\int_0^t \frac{1}{4+\cos(2u)}\:du$$
Then double angle identity on the last integral
$$\int_0^t \frac{1}{5\cos^2u + 3\sin^2u}\:du = \frac{1}{5}\int_0^t \frac{\sec^2u}{1+\frac{3}{5}\tan^2 u}\:du = \frac{1}{\sqrt{15}}\tan^{-1}\left(\sqrt{\frac{3}{5}}\tan t\right)$$
But since the antiderivative needs to be continuous, we need to add a staircase term to raise up the discontinuous points of the function to appropriately connect them. Then adding up all of the results gets us
$$y(t) = \cos(2t) + t\sin(2t) - \frac{4}{\sqrt{15}}\sin(2t)\tan^{-1}\left(\sqrt{\frac{3}{5}}\tan t\right)$$ $$-\frac{4\pi}{\sqrt{15}}\sin(2t)\left\lfloor \frac{1}{\pi}t+\frac{1}{2}\right\rfloor+\frac{1}{2}\cos(2t)\log(4+\cos(2t))$$
This behavior makes perfect sense since the driving force (inhomogeneous term) is in resonance with the homogeneous solution, leading to the growing amplitude term.
Best Answer
Taking the Laplace transform of the left hand side, using the convolution formula on the integral (note the integral is $f\star g$ where $f $ is the identity function and and $g=x$): $${\cal L} \Bigl( x'(t)+\int^{t}_{0}(t-s)x(s)ds\Bigr) = s X(s)-x(0) + {1\over s^2}\cdot X(s). $$
Taking the Laplace transform of the right hand side: $$ {\cal L}\Bigl( t+\frac{1}{2}t^2+\frac{1}{24}t^4 \Bigr)= {1\over s^2}+{1\over 2}{2\over s^3}+{1\over 24}\cdot{24\over s^5}. $$ So, we have: $$ s X(s)-x(0) + {1\over s^2}\cdot X(s)= {1\over s^2}+ {1\over s^3}+ {1\over s^5}. $$
Now solve for $X(s)$ and then take the inverse transform to find $x(t)$.