Question

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

Answer 1

The integral does not converge.

Explanation:

Notice that `e^(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.

`inte^(3/x)/x^2dx`

Let `u=3/x` so `du=-3/x^2dx`.

`=-1/3inte^(3/x)(-3/x^2dx)=-1/3inte^udu=-1/3e^u=-1/3e^(3/x)`

So:

`int_0^1e^(3/x)/x^2dx=lim_(brarr0^+)int_b^1e^(3/x)/x^2dx=lim_(brarr0^+)[-1/3e^(3/x)]_b^1`

Evaluating:

`=-1/3e^3+lim_(brarr0^+)1/3e^(3/b)`

As `brarr0` from the right, we see that `3/brarroo`. Thus `lim_(brarr0^+)1/3e^(3/b)=oo` as well and the integral will not converge.