TITLE: $\int_3^5{\frac{x^2}{1+x^2}dx}$ by differentiation under the integral
QUESTION [7 upvotes]: 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.

REPLY [2 votes]: Let's observe the next differential equation:
$y''-y+ \frac{\cos3x-\cos5x}{x}=0$ ,
This is nonhomogeneous second-order ordinary differential equation of the form:
$y''+p(x)y'+q(x)y-g(x)=0$ ,which can be solved if the general solution to the homogenous version is known, in which case variation of parameters can be used to find the particular solution:
$y_p=-y_1(x)\int\frac{y_2(x)g(x)}{W(x)}\, dx + y_2(x)\int\frac{y_1(x)g(x)}{W(x)}\, dx$ , where $y_p$ is particular solution, $y_1(x)$ and $y_2(x)$ are the homogeneous solutions of the equation $y''+p(x)y'+q(x)y=0$ and  W(x) is the Wronskian of these two functions.