I want to find the: $$\int_2^3{\frac{e^x}{x^2}}\text{d}x$$ So I am thinking of using the integration by parts method. I have:
$$\int_2^3{\frac{e^x}{x^2}}\text{ d}x=\int_2^3{e^x\cdot{x^{-2}}}\text{ d}x=\int_2^3{(e^x)'\cdot{x^{-2}}}\text{ d}x$$ But I don't know how to continue, because in my understanding I need somehow to appear the first integral. Any ideas?
On
$$ \int_{2}^{3}\frac{e^x}{x^2}\,dx = \int_{1/3}^{1/2} e^{1/t}\,dt=\int_{1/3}^{1/2}\sum_{k\geq 0}\frac{1}{k! t^k}\,dt=\frac{1}{6}+\log\frac{3}{2}+\sum_{n\geq 1}\frac{3^n-2^n}{n\cdot(n+1)!} $$ is not an elementary integral (it depends on the exponential integral), but the numerical evaluation of the last series is not a huge challenge. Simple bounds can be derived through the Cauchy-Schwarz or Hermite-Hadamard inequalities, given the convexity of $e^{1/t}$ over $\left[\frac{1}{2},\frac{1}{3}\right]$. By Simpson's rule $$ \int_{2}^{3}\frac{e^x}{x^2}\,dx \stackrel{\text{IBP}}{=} -\frac{e^3}{3}+\frac{e^2}{2}+\int_{2}^{3}\frac{e^x}{x}\,dx\approx -\frac{e^3}{3}+\frac{e^2}{2}+\frac{e^3}{6}\left(\frac{1}{3}+\frac{8}{5\sqrt{e}}+\frac{1}{2e}\right)\approx 2. $$
Note that a step of integration by parts should be
$$\int {\frac{e^x}{x^2}}\text{d}x= \frac{e^x}{x^2}-\int {-\frac{e^x}{x}}\text{d}x$$
but for $\int {\frac{e^x}{x}}\text{d}x$ we need to refer to special functions.