### Video Transcript

Use the integral test to determine whether the series the sum from π equals one to β of two π plus one divided by three π squared plus three π plus one is convergent or divergent.

In this question, weβre given an infinite series. And we need to determine the convergence or divergence of this series by using the integral test. So letβs start by recalling what the integral test tells us. This says if π of π₯ is a continuous, positive, decreasing function for all values of π₯ greater than or equal to some nonnegative integer π and π evaluated at π is equal to the sequence ππ for all of our values of π, then the integral test tells us the following. If the integral from π to β of π of π₯ with respect to π₯ is convergent, then our series the sum from π equals π to β of ππ is also convergent. However, if the integral from π to β of π of π₯ with respect to π₯ is divergent, then our series the sum from π equals π to β of ππ is divergent.

So the integral test is a method of turning a question about the convergence or divergence of a series into a question about the convergence or divergence of an integral. Letβs try applying this to the series given to us in the question. First, we can see our series starts from π is equal to one. So weβll set our value of π equal to one. And in fact, weβll update our integral test with the value of π set to be one. Next, remember, π evaluated at π is supposed to be equal to our summand ππ, so weβll set π of π₯ to be equal to our summand with π₯ instead of π.

We then need to show that this function π of π₯ is continuous, positive, and decreasing for all values of π₯ greater than or equal to one. Letβs start with showing this is a continuous function. First, we see π of π₯ is equal to two π₯ plus one all divided by three π₯ squared plus three π₯ plus one. We can see this is the quotient of two polynomials. So π of π₯ is a rational function. And we know rational functions are continuous across their entire domain. In fact, for rational functions, the only time they wonβt be continuous is when the denominator is equal to zero. So we need to check where our denominator is equal to zero.

This means we would need to factor our quadratic. However, if we tried to do this, we would see we get no real roots. For example, our discriminant π squared minus four ππ gives us three squared minus four times three times one, which is nine minus 12, which is equal to negative three. And because this is negative, that means our quadratic has no real roots. So our denominator is never equal to zero. And therefore, our function π of π₯ is defined for real values of π₯, which means itβs also continuous for all real values of π₯.

Letβs move on to showing that π of π₯ is a positive function for all values of π₯ greater than or equal to one. To do this, letβs consider what happens when we substitute in a value of π₯ is greater than or equal to one. In fact, by doing this, we can see every term in this expression will be positive. So when π₯ is greater than or equal to one, π of π₯ is the quotient to two positive numbers, which means itβs positive.

Finally, we need to show that π of π₯ is decreasing for all values of π₯ greater than or equal to one. Since π of π₯ is a rational function, weβll do this by using the quotient rule. Remember π of π₯ will be decreasing on any intervals where its slope is negative. So by using the quotient rule, we get that π prime of π₯ is equal to the derivative of two π₯ plus one with respect to π₯ times three π₯ squared plus three π₯ plus one minus the derivative of three π₯ squared plus three π₯ plus one with respect to π₯ times two π₯ plus one all divided by three π₯ squared plus three π₯ plus one all squared.

Now, we can simplify this expression by evaluating our derivatives by using the power rule for the differentiation. First, the derivative of two π₯ plus one with respect to π₯ is equal to two. Next, the derivative of three π₯ squared plus three π₯ plus one is equal to six π₯ plus three. So π prime of π₯ simplifies to give us the following expression. And we can simplify this by distributing two over our first set of parentheses and multiplying the two sets of parentheses in our numerator. Doing this gives us six π₯ squared plus six π₯ plus two minus 12π₯ squared plus 12π₯ plus three all divided by three π₯ squared plus three π₯ plus one all squared.

Now, if we distribute the negative one over our parentheses in our numerator and simplify, we get negative six π₯ squared minus six π₯ minus one all divided by three π₯ squared plus three π₯ plus one. Now, we can evaluate the slope of our function π of π₯ when π₯ is greater than or equal to one. When π₯ is greater than or equal to one, negative six π₯ squared is negative, negative six π₯ is negative, and negative one is negative. So our numerator will always be negative. However, our denominator is the square of a nonzero number. So our denominator will always be positive. Therefore, when π₯ is greater than or equal to one, our slope function will be the quotient between a negative number and a positive number, so it will always be negative. And if the slope of our function is always negative when π₯ is greater than or equal to one, then it must be decreasing on this interval.

So now, weβve shown all our prerequisites for using the integral test are true. This means we can look at the convergence or divergence of the integral from one to β of two π₯ plus one divided by three π₯ squared plus three π₯ plus one with respect to π₯ to determine the convergence or divergence of our series. To do this, weβll start by using our laws of improper integrals. We know the convergence or divergence of this integral is the same as the convergence or divergence of the limit as π‘ approaches β of the integral from one to π‘ of two π₯ plus one divided by three π₯ squared plus three π₯ plus one with respect to π₯.

And now our integrand is continuous on our interval of integration, so we can use any of our tools for evaluating definite integrals. And the easiest way to evaluate this integral is to notice something interesting about our integrand. If we differentiate our denominator with respect to π₯, we get six π₯ plus three. And we can see that our numerator is a linear multiple of six π₯ plus three. If we multiply our numerator by three and divide our entire integral by three, we get a special form for our integral. Our integrand is in the form π prime of π₯ divided by π of π₯. And we know the integral of π prime of π₯ divided by π of π₯ with respect to π₯ is equal to the natural logarithm of the absolute value of π of π₯ plus a constant of integration πΆ.

So by using this integral rule and ignoring our constant of integration β because weβre evaluating a definite integral β we get the limit as π‘ approaches β of one-third times the natural logarithm of the absolute value of three π₯ squared plus three π₯ plus one evaluated at the limits of integration π‘ and one. Before we evaluate this at the limits of integration, weβll take the constant factor of one-third outside of our limit. Then weβll just evaluate this at the limits of integration.

We get one-third times the limit at π‘ approaches β of the natural logarithms of the absolute value of three π‘ squared plus three π‘ plus one minus the natural logarithm of the absolute value of three times one squared plus three times one plus one. And now, we can start simplifying. First, three times one squared plus three times one plus one is equal to seven. And we could keep simplifying here. However, at this point, we can just evaluate our limit directly. We see our limit is as π‘ approaches β. As π‘ approaches β, the quadratic three π‘ squared plus three π‘ plus one is approaching β. This means taking the absolute value of this expression will still approach β. And of course, we know the natural logarithm of π₯ approaches β as π₯ approaches β. So in this case, it will also approach β. And the natural logarithm of the absolute value of seven is a constant. So it doesnβt change the value as π‘ changes.

Therefore, this limit must be divergent. But remember this limit being divergent means that our integral is divergent. And by the integral test, this integral being divergent means that our series is divergent. Therefore, by using the integral test, we were able to determine the sum from π equals one to β of two π plus one all divided by three π squared plus three π plus one is divergent.