Example: Evaluate (if it converges)

Solution: There is one obvious problem, namely the negative infinity. We check the denominator and find two troublesome points. One of them, x = 1, is not in the integration interval and we can ignore it. The second root is −1 and it is the second problem in our integral. Therefore we have to split the given integral into more integrals of the basic improper type. We start by integrating from negative infinity, but since this is already a problem, we have to stop before we get to −1. For instance, −2 seems a nice number to stop. The second integral will go from −2 to −1 (one problem at the upper limit) and the third integral will go from −1 to 0 (one problem at the lower limit).

Now we find an antiderivative using the partial fractions decomposition.

Note how we put the logarithms together so that we can apply limits to just one expression. Now we evaluate the three integrals, avoiding problem points using limits:

So the first one is convergent.

This integral is divergent and therefore the given integral is divergent, the third part won't change this.

Does the given integral exist? For that we need to evaluate also the third part.

We compare all parts and see that have opposite infinities, so the given integral does not even exist.

Notes:
1. It can be proved that the negative infinity is of the same "size" as the positive infinity of the second integral. However, as we remarked, these infinities do not cancel in the sense that they would make the whole integral convergent. Once a part of region is funny, the whole region is funny.

2. One frequent mistake is that people forget to check for problems inside the integration interval. In this example it would lead to the impression that the given integral itself is of the basic type and we can directly calculate:

As we now know, this answer is incorrect. Note that if a student makes this erroneous calculation, there is no way to see from the result that it is wrong. Since the integrated function is both positive and negative, it could well be that the parts above and below the x-axis have convergent areas of equal size, therefore they cancel. It is really important to check all definite integrals for troubles before one starts evaluating them.

3. If one neglects to put the logarithms together, it quickly leads to trouble:

As we remarked before, this trouble is of our own doing and has nothing to do with convergence of the integral.