I'm trying an easy problem to get my bearings using the method here.

The integral is $$\int_3^5{\frac{x^2}{1+x^2}dx}$$.

I would like to proceed, if possible to solve by defining:

$$F(y) = \int_3^5{\frac{\sin{(y\cdot x})}{1+x^2}dx}$$

Then obtaining $$-F''(y) = \int_3^5{\frac{x^2\sin{(y\cdot x})}{1+x^2}dx}$$

Adding gives $$F(y) - F''(y)= \int_3^5{\frac{(1+x^2)\sin{(y\cdot x})}{1+x^2}dx}$$

or $$F(y) - F''(y) - \frac{\cos{3y}-\cos{5y}}{y} = 0.$$

This is where I'm stuck. I don't know how to solve the differential equation. Any help would be greatly appreciated. I'm assuming that this can be done.