Given: $\displaystyle f(x) = \int_x^{x+1} \sin e^t dt$. Show: $\displaystyle e^x f(x) = \cos (e^x) - e^{-1} \cos (e^{x+1}) + r(x),$ where $\displaystyle |r(x)| < Ce^{-x}, C \in \mathbb{R}.$

I was fairly easily able to prove that $\displaystyle $e^x f(x) = \cos (e^x) - e^{-1} \cos (e^{x+1}) - e^x \int_{e^x}^{e^{x+1}} \frac{\cos u}{u^2} du,$$ so I won't go through all those steps.

Where I'm having trouble is the last part, namely, showing that $\displaystyle | $e^x \int_{e^x}^{e^{x+1}} \frac{\cos u}{u^2} du| < Ce^{-x}$$. How do I do this? I tried $\displaystyle | $\int_{e^x}^{e^{x+1}} \frac{\cos u}{u^2} du| < |\int_{e^x}^{e^{x+1}} \frac{du}{u^2} |$$, but that did not produce the desired result.