calculus - How to solve this problem on the differentiation under integral sign

I was solving problems on differentiation under integral sign :

I encountered this problem :
Evaluate this integral $$I=\int_0^\infty\frac{(e^{-ax}-e^{-4bx})sin(x)}{x}dx$$
and then use it to solve the integral
$$I_1=\int_0^\infty\frac{(x^3e^{-2x}-x^2e^{-3x}+1)sin(x)}{x}dx$$

I evaluated the integral I , then I divided integral I1 to 3 integrals , I can solve the first 2 integrals (by differentiating the integral as I clarify below).. My problem is in the integration of sin(x) over x .. How can I deduce it from the integral I ?
$$I_1=\int_0^\infty x^2e^{-2x}sin(x)dx+\int_0^\infty-x e^{-3x}sin(x)dx+\int_0^\infty \frac{sin(x)}{x}dx$$

Here is my solution to evaluate I , then the first 2 parts of the integral I1: ( I know the below solution is correct .. But how to complete to evaluate the last part of I1)
$$\frac{\partial I}{\partial a}=\int_0^\infty - e^{-ax} sin(x)=\frac{-1}{1+a^2}$$
$$I=\int\frac{-1}{1+a^2}da+f(b)$$
$$I=-\tan^{-1}(a)+f(b)$$
$$f(b)=\tan^{-1}(4b)$$
$$I=-\tan^{-1}(a)+\tan^{-1}(4b)$$
$$\frac{\partial^2 I}{\partial a^2}=\int_0^\infty x e^{-ax} sin(x)=\frac{d}{da}(\frac{-1}{1+a^2})=\frac{2a}{(1+a^2)^2}$$

$$\frac{\partial^3 I}{\partial a^3}=\int_0^\infty -x^2 e^{-ax} sin(x)=\frac{d}{da}(\frac{2a}{(1+a^2)^2})=\frac{-8a^3+2a^2-8a+2}{(a^2+1)^3}$$
$$\int_0^\infty x^2e^{-2x}sin(x)dx=-\frac{\partial^3 I}{\partial a^3}|_{a=2}$$
$$\int_0^\infty-x e^{-3x}sin(x)dx=-\frac{\partial^2 I}{\partial a^2}|_{a=3}$$