# How do you integrate int e^(3/x)/x^2dx from [0,1]?

Jan 14, 2017

The integral does not converge.

#### Explanation:

Notice that ${e}^{\frac{3}{x}} / {x}^{2}$ is undefined at $x = 0$, so we are working with an improper integral. Instead of having a bound of $0$, we will take the limit of the integral as some variable approaches $0$.

Before doing so, first find the general antiderivative of the function.

$\int {e}^{\frac{3}{x}} / {x}^{2} \mathrm{dx}$

Let $u = \frac{3}{x}$ so $\mathrm{du} = - \frac{3}{x} ^ 2 \mathrm{dx}$.

$= - \frac{1}{3} \int {e}^{\frac{3}{x}} \left(- \frac{3}{x} ^ 2 \mathrm{dx}\right) = - \frac{1}{3} \int {e}^{u} \mathrm{du} = - \frac{1}{3} {e}^{u} = - \frac{1}{3} {e}^{\frac{3}{x}}$

So:

${\int}_{0}^{1} {e}^{\frac{3}{x}} / {x}^{2} \mathrm{dx} = {\lim}_{b \rightarrow {0}^{+}} {\int}_{b}^{1} {e}^{\frac{3}{x}} / {x}^{2} \mathrm{dx} = {\lim}_{b \rightarrow {0}^{+}} {\left[- \frac{1}{3} {e}^{\frac{3}{x}}\right]}_{b}^{1}$

Evaluating:

$= - \frac{1}{3} {e}^{3} + {\lim}_{b \rightarrow {0}^{+}} \frac{1}{3} {e}^{\frac{3}{b}}$

As $b \rightarrow 0$ from the right, we see that $\frac{3}{b} \rightarrow \infty$. Thus ${\lim}_{b \rightarrow {0}^{+}} \frac{1}{3} {e}^{\frac{3}{b}} = \infty$ as well and the integral will not converge.