I'd like to get some feedback on an integral calculation, if anyone might spot something wrong with my work:
$$\int_4^9{\frac{2e^{\sqrt{x}}}{\sqrt{x}}}\,\,dx$$
Using substitution, let
$$u = \sqrt{x},\,\,\frac{du}{dx}=\frac{1}{2\sqrt{x}},\,\,du=\frac{1}{2\sqrt{x}}dx$$
Working out the indefinite integral
$$\int{\frac{2e^{\sqrt{x}}}{\sqrt{x}}}\,\,dx = 4\int{e^u}\,du = 4\cdot e^u+c = 4 \cdot e^{\sqrt{x}}+c$$
Calculating the definite integral bounds
$$= 4\cdot e^{\sqrt{9}} - 4\cdot e^{\sqrt{4}} = 4\cdot e^3 - 4\cdot e^2 = 4\cdot e^2(e - 1)$$
I'd appreciate any feedback on this calculation! Thanks.
Perhaps easier, "identifying" derivatives of functions:
$$\int_4^9\frac{2e^{\sqrt x}}{\sqrt x}dx=4\int_4^9e^{\sqrt x}\;(\sqrt x)' dx=\left.4e^{\sqrt x}\right|_4^9=4(e^3-e^2)$$