Note that : $\int_{0}^{\infty}\frac{cos^{2}(x)}{1+x^2} = 0.5\int_{0}^{\infty}\frac{1+2cos(2x)}{1+x^2} = \frac{\pi}{4} + \int_{0}^{\infty}\frac{cos(2x)}{1+x^{2}} = \frac{\pi}{4}+\frac{\pi}{2}e^{-2}$. The last one is Laplace's integral.
For second part : $\displaystyle \int_{0}^{\infty} e^{-(x^{2}+1)y}dy = \frac{1}{1+x^2}$, so you should consider $\displaystyle \int_{0}^{\infty}dy \int_{0}^{\infty}e^{-(x^{2}+1)y}\cos(ax)dx$. Now it will be easy to calculate Laplace's integral.(just find $\int_{0}^{\infty} e^{ax}\cos(bx)$ using integration by parts).
Note that : $\int_{0}^{\infty}\frac{cos^{2}(x)}{1+x^2} = 0.5\int_{0}^{\infty}\frac{1+2cos(2x)}{1+x^2} = \frac{\pi}{4} + \int_{0}^{\infty}\frac{cos(2x)}{1+x^{2}} = \frac{\pi}{4}+\frac{\pi}{2}e^{-2}$. The last one is Laplace's integral.
For second part : $\displaystyle \int_{0}^{\infty} e^{-(x^{2}+1)y}dy = \frac{1}{1+x^2}$, so you should consider $\displaystyle \int_{0}^{\infty}dy \int_{0}^{\infty}e^{-(x^{2}+1)y}\cos(ax)dx$. Now it will be easy to calculate Laplace's integral.(just find $\int_{0}^{\infty} e^{ax}\cos(bx)$ using integration by parts).