Derivative of integral with infinity as upper bound

Question

What is the solution to the derivative of following integral? I know how to take derivatives of integrals but I never came across one with infinity in one of his bounds.

$F(t) = \int^{\infty}_t \frac{x-4}{(x^2+4)(x+1)}$

$t >= 0$

Answer 1

Hint

If you want to go back to a situation you already know, just change varieble $x=\frac 1y$ and so $$I = \int_a^b\frac{x-4}{(x^2+4)(x+1)}dx=\int_{\frac 1a}^{\frac 1b}\frac{4 y-1}{(y+1) \left(4 y^2+1\right)}dy=-\int_{\frac 1b}^{\frac 1a}\frac{4 y-1}{(y+1) \left(4 y^2+1\right)}dy$$

Now, what are the bounds in your case ?

In any manner, what David Mitra suggested is really the simplest solution.

I am sure that you can take from here.