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### Section 1-8 : Improper Integrals

9. Determine if the following integral converges or diverges. If the integral converges determine its value.

\[\int_{1}^{4}{{\frac{1}{{{x^2} + x - 6}}\,dx}}\]Show All Steps Hide All Steps

First, notice that there is a division by zero issue (and hence a discontinuity) in the integrand at \(x = 2\) and note that this is between the limits of the integral. We know that as long as that discontinuity is there we can’t do the integral.

However, recall from the notes in this section that we can only deal with discontinuities that if they occur at one of the limits of the integral. So, we’ll need to break up the integral at \(x = 2\) .

\[\int_{1}^{4}{{\frac{1}{{{x^2} + x - 6}}\,dx}} = \int_{1}^{2}{{\frac{1}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\,dx}} + \int_{2}^{4}{{\frac{1}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\,dx}}\]Remember as well, that we can only break up the integral like this provided both of the new integrals are convergent! If it turns out that even one of them is divergent then it will turn out that we couldn’t have done this and the original integral will be divergent.

So, not worrying about if this was really possible to do or not, let’s proceed with the problem.

We can eliminate the discontinuity in each as follows,

\[\int_{1}^{4}{{\frac{1}{{{x^2} + x - 6}}\,dx}} = \mathop {\lim }\limits_{t \to \,{2^ - }} \int_{1}^{t}{{\frac{1}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\,dx}} + \mathop {\lim }\limits_{s \to \,{2^ + }} \int_{s}^{4}{{\frac{1}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\,dx}}\]Don’t forget that the limits on these kinds of integrals must be one-sided limits.

Show Step 2Next, let’s do the integral. We’ll not be putting a lot of explanation/detail into the integration process. By this point it is assumed that your integration skills are getting pretty good. If you find your integration skills are a little rusty you should go back and do some practice problems from the appropriate earlier sections.

In this case we will need to do some partial fractions in order to the integral. Here is the partial fraction work.

\[\frac{1}{{\left( {x + 3} \right)\left( {x - 2} \right)}} = \frac{A}{{x + 3}} + \frac{B}{{x - 2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,1 = A\left( {x - 2} \right) + B\left( {x + 3} \right)\] \[\begin{array}{ll}{x = 2:} & {1 = 5B}\\ & \\ {x = - 3:} & {1 = - 5A}\end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\begin{array}{l}{\displaystyle A = - \frac{1}{5}}\\{\displaystyle B = \frac{1}{5}}\end{array}\]The integration work is then,

\[\int{{\frac{1}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\,dx}} = \int{{\frac{{\frac{1}{5}}}{{x - 2}} - \frac{{\frac{1}{5}}}{{x + 3}}\,dx}} = \frac{1}{5}\ln \left| {x - 2} \right| - \frac{1}{5}\ln \left| {x + 3} \right| + c\]Note that we didn’t do the definite integral here. The limits don’t really affect how we do the integral and the integral for each was the same with only the limits being different so no reason to do the integral twice.

Show Step 3Okay, now let’s take care of the limits on the integral.

\[\begin{align*}\int_{1}^{4}{{\frac{1}{{{x^2} + x - 6}}\,dx}} & = \mathop {\lim }\limits_{t \to \,{2^ - }} \left. {\left( {\frac{1}{5}\ln \left| {x - 2} \right| - \frac{1}{5}\ln \left| {x + 3} \right|} \right)} \right|_1^t + \mathop {\lim }\limits_{s \to \,{2^ + }} \left. {\left( {\frac{1}{5}\ln \left| {x - 2} \right| - \frac{1}{5}\ln \left| {x + 3} \right|} \right)} \right|_s^4\\ & = \mathop {\lim }\limits_{t \to \,{2^ - }} \left( {\frac{1}{5}\ln \left| {t - 2} \right| - \frac{1}{5}\ln \left| {t + 3} \right| - \left( {\frac{1}{5}\ln \left( 1 \right) - \frac{1}{5}\ln \left( 4 \right)} \right)} \right)\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \mathop {\lim }\limits_{s \to \,{2^ + }} \left( {\frac{1}{5}\ln \left( 2 \right) - \frac{1}{5}\ln \left( 7 \right) - \left( {\frac{1}{5}\ln \left| {s - 2} \right| - \frac{1}{5}\ln \left| {s + 3} \right|} \right)} \right)\end{align*}\] Show Step 4We now need to evaluate the limits in our answer from the previous step. Here is the limit work.

\[\begin{align*}\int_{1}^{4}{{\frac{1}{{{x^2} + x - 6}}\,dx}} & = \mathop {\lim }\limits_{t \to \,{2^ - }} \left( {\frac{1}{5}\ln \left| {t - 2} \right| - \frac{1}{5}\ln \left| {t + 3} \right| - \left( {\frac{1}{5}\ln \left( 1 \right) - \frac{1}{5}\ln \left( 4 \right)} \right)} \right)\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \mathop {\lim }\limits_{s \to \,{2^ + }} \left( {\frac{1}{5}\ln \left( 2 \right) - \frac{1}{5}\ln \left( 7 \right) - \left( {\frac{1}{5}\ln \left| {s - 2} \right| - \frac{1}{5}\ln \left| {s + 3} \right|} \right)} \right)\\ & = \,\,\,\,\,\,\,\,\,\left[ { - \infty - \frac{1}{5}\ln \left( 5 \right) + \frac{1}{5}\ln \left( 4 \right)} \right]\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,\left[ {\frac{1}{5}\ln \left( 2 \right) - \frac{1}{5}\ln \left( 7 \right) + \frac{1}{5}\ln \left( 5 \right) + \infty } \right]\end{align*}\]Note that we put the answers for each limit in brackets to make it clear what each limit was. This will be important for the next step.

Show Step 5The final step is to write down the answer!

Now, from the limit work in the previous step we see that,

\[\begin{align*}\int_{1}^{2}{{\frac{1}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\,dx}} & = \mathop {\lim }\limits_{t \to \,{2^ - }} \left( {\frac{1}{5}\ln \left| {t - 2} \right| - \frac{1}{5}\ln \left| {t + 3} \right| - \left( {\frac{1}{5}\ln \left( 1 \right) - \frac{1}{5}\ln \left( 4 \right)} \right)} \right) = \, - \infty \\ & \\ \int_{2}^{4}{{\frac{1}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\,dx}} & = \mathop {\lim }\limits_{s \to \,{2^ + }} \left( {\frac{1}{5}\ln \left( 2 \right) - \frac{1}{5}\ln \left( 7 \right) - \left( {\frac{1}{5}\ln \left| {s - 2} \right| - \frac{1}{5}\ln \left| {s + 3} \right|} \right)} \right) = \infty \end{align*}\]Therefore, each of these integrals are divergent. This means that we were, in fact, not able to break up the integral as we did back in Step 1.

This in turn means that the integral **diverges**.