Lesson Video: Integrals Resulting in Logarithmic Functions Mathematics • Third Year of Secondary School

Video Transcript

Integrals involving logarithmic functions. In this video, we will learn how to use integration by substitution on function of the form 𝑓 prime of 𝑥 over 𝑓 of 𝑥. We’ll be looking at some examples to see the kind of integrals we can use this method on. Now, let’s start by considering the following integral. And that’s the indefinite integral of 𝑓 prime of 𝑥 over 𝑓 of 𝑥 with respect to 𝑥. And we’re going to try to solve this integral. And we can in fact do this with a substitution. Let’s let 𝑢 equal 𝑓 of 𝑥. And we can differentiate 𝑢 with respect to 𝑥. And this gives us d𝑢 by d𝑥 is equal to 𝑓 prime of 𝑥. Where the prime denotes a differentiation with respect to 𝑥. And this tells us that d𝑢 is equal to 𝑓 prime of 𝑥 d𝑥.

Now we can rewrite our integral as the integral of one over 𝑓 of 𝑥 multiplied by 𝑓 prime of 𝑥 d𝑥. We’re now ready to perform our substitution. We will substitute in 𝑢 for 𝑓 of 𝑥 and d𝑢 for 𝑓 prime of 𝑥 d𝑥. So you obtain that our integral is equal to the integral of one over 𝑢 with respect to 𝑢. And this is an integral which we know how to solve. We know that the integral of one over 𝑥 with respect to 𝑥 is equal to the natural logarithm of the absolute value of 𝑥 plus 𝑐. Therefore, our integral is equal to the natural logarithm of the absolute value of 𝑢 plus our integration constant 𝑐. And here, we can substitute our value of 𝑢 back into our equation. And this leads us to our result, which is that the indefinite integral of 𝑓 prime of 𝑥 over 𝑓 of 𝑥 with respect to 𝑥 is equal to the natural logarithm of the absolute value of 𝑓 of 𝑥 plus 𝑐.

Let’s now look at an example of how this works.

Determine the indefinite integral of two 𝑥 plus one over 𝑥 squared plus 𝑥 minus seven with respect to 𝑥.

When we look at the integrand of this integral, we can notice that the numerator looks a lot like the derivative of the denominator. Let’s quickly check this. We can call the denominator 𝑓 of 𝑥. So 𝑓 of 𝑥 is equal to 𝑥 squared plus 𝑥 minus seven. And now we can differentiate this. Using the power rule on each term, we obtain that 𝑓 prime of 𝑥 is equal to two 𝑥 plus one. Which is equal to the numerator of our fraction. Therefore, our integrand is of the form 𝑓 prime of 𝑥 over 𝑓 of 𝑥. And we in fact have a rule for integrating functions of this form. It tells us that the integral of 𝑓 prime of 𝑥 over 𝑓 of 𝑥 with respect to 𝑥 is equal to the natural logarithm of the absolute value of 𝑓 of 𝑥 plus 𝑐. Using this rule, we obtain that the integral of two 𝑥 plus one over 𝑥 squared plus 𝑥 minus seven with respect to 𝑥 is equal to the natural logarithm of the absolute value of 𝑥 squared plus 𝑥 minus seven plus our constant of integration 𝑐.

In the next example, we will see how this method can be very useful when integrating certain trigonometric functions.

Determine the indefinite integral of cot 𝑥 with respect to 𝑥.

Now we know that cot 𝑥 can be written as cos of 𝑥 over sin of 𝑥. Therefore, we can rewrite our integral as the integral of cos 𝑥 over sin 𝑥 with respect to 𝑥. Next, we’ll be using the fact that the differential of sin 𝑥 with respect to 𝑥 is equal to cos 𝑥. And so if we let our denominator sin 𝑥 be equal to 𝑓 of 𝑥, then our numerator, cos of 𝑥, will be equal to 𝑓 prime of 𝑥. And here, we can see that our integral is of the form of the integral of 𝑓 prime of 𝑥 over 𝑓 of 𝑥 with respect to 𝑥.

And we know a formula for solving integrals of this form. And this formula tells us that the indefinite integral of 𝑓 prime of 𝑥 over 𝑓 of 𝑥 with respect to 𝑥 is equal to the natural logarithm of the absolute value of 𝑓 of 𝑥 plus 𝑐. If we apply this formula to our integral with 𝑓 of 𝑥 is equal to sin 𝑥, then we will reach our solution. Which is that the indefinite integral of cot 𝑥 with respect to 𝑥 is equal to the natural logarithm of the absolute value of sin 𝑥 plus 𝑐.

Now we can in fact take this method one step further. And consider when our numerator differs from the differential of the denominator by a constant factor. So if our integrand is equal to 𝑎 times 𝑓 prime of 𝑥 over 𝑓 of 𝑥, where 𝑎 is a real number. Now, this may appear very straightforward. Since we can use the fact that we can factor out a constant from inside an integral. This gives us that our integral is equal to 𝑎 multiplied by the natural logarithm of the absolute value of 𝑓 of 𝑥 plus 𝑐. And now this method seems very straightforward. It’s just a slight step up from what we we’re doing before. However, there are in fact two potentially quite difficult parts in setting up our integral. So we can apply this formula.

And the first of these two points is to actually spot that our integral can be in this form. Since it’s not always obvious that our numerator is some constant multiplied by the differential of the denominator. Sometimes we may be required to multiply our integrand by some function over itself, such as ℎ of 𝑥 over ℎ of 𝑥. In order to get it in the form to use this method. The second part, which is often slightly less tricky, is to find the value of our constant 𝑎, since it’s not always immediately obvious.

Let’s now look at an example of how we can use this method.

Determine the indefinite integral of 𝑥 squared plus seven over 𝑥 cubed plus 21𝑥 minus five with respect to 𝑥.

Now the first thing that we notice is that the numerator of our integrand looks a lot like the differential of the denominator. Let’s quickly check this. Let’s let 𝑓 of 𝑥 be our denominator. So 𝑓 of 𝑥 is equal to 𝑥 cubed plus 21𝑥 minus five. And now we differentiate. And we get that 𝑓 prime of 𝑥 is equal to three 𝑥 squared plus 21, which is not equal to our numerator. However, we notice that it’s in fact out by a factor of three. If we divide 𝑓 prime of 𝑥 by three, then we obtain that one-third 𝑓 prime of 𝑥 is equal to 𝑥 squared plus seven. And this is now equal to the numerator of our integrand. We could write our integral in terms of 𝑓 of 𝑥. And we would obtain that it is equal to the integral of one-third 𝑓 prime of 𝑥 over 𝑓 of 𝑥 with respect to 𝑥.

Now we can use a rule which we know to integrate integrals of this form. It tells us that the integral of 𝑎 multiplied by 𝑓 prime of 𝑥 over 𝑓 of 𝑥 with respect to 𝑥 is equal to 𝑎 multiplied by the natural logarithm of the absolute value of 𝑓 of 𝑥 plus 𝑐. So we can apply this formula. Except in our case, 𝑎 is one-third. And 𝑓 of 𝑥 is equal to 𝑥 cubed plus 21𝑥 minus five. In doing this, we reach our solution, which is that the indefinite integral of 𝑥 squared plus seven over 𝑥 cubed plus 21𝑥 minus five with respect to 𝑥 is equal to one-third multiplied by the natural logarithm of the absolute value of 𝑥 cubed plus 21𝑥 minus five plus 𝑐.

In this last example, we saw how we can find our constant 𝑎 by inspection. However, this is not always particularly obvious. Another way in which we can find it is by using substitution as we’ll see in the next example.

Determine the indefinite integral of 27 sin 𝑥 plus 21 cos 𝑥 all over seven sin 𝑥 minus nine cos 𝑥 with respect to 𝑥.

Here, we notice that the numerator of the integrand looks as though it could be the differential of the denominator. Since the differential of sin 𝑥 with respect to 𝑥 is equal to cos 𝑥 and the differential of negative cos 𝑥 with respect to 𝑥 is equal to sin 𝑥. Now it looks as though the numerator may differ from the differential of the denominator by a constant factor. However, we do not know what this factor is. We can try and find it by using a substitution. Let’s let 𝑢 be equal to the denominator of the integrand, so that seven sin 𝑥 minus nine cos 𝑥. Now we can differentiate 𝑢 with respect to 𝑥. Using the fact that sine differentiates to cos 𝑥 and negative cos 𝑥 differentiates to sin 𝑥. We obtain that d𝑢 by d𝑥 is equal to seven cos 𝑥 plus nine sin 𝑥. This gives us that d𝑢 is equal to nine sin 𝑥 plus seven cos 𝑥 d𝑥.

Now let’s rearrange our integral so we can apply this substitution. We notice that we can factor out a factor of three from our numerator. And this enables us to write our integral as the integral of three over seven sin 𝑥 minus nine cos 𝑥 multiplied by nine sin 𝑥 plus seven cos 𝑥 d𝑥. And so we can substitute 𝑢 into the denominator of our fraction. And we can substitute in d𝑢 for nine sin 𝑥 plus seven cos 𝑥 d𝑥. Giving us that it is equal to the integral of three over 𝑢 d𝑢, which can be integrated to three multiplied by the natural logarithm of the absolute value of 𝑢 plus 𝑐. For our final step, we simply substitute back in seven sin 𝑥 minus nine cos 𝑥 for 𝑢. This gives us our solution, which is three multiplied by the natural logarithm of the absolute value of seven sin 𝑥 minus nine cos 𝑥 plus 𝑐.

Now this method of integration can be used to integrate many different types of functions. In the next example, we’ll see an integration of a function involving a natural logarithm.

Determine the indefinite integral of negative three over 𝑥 multiplied by the natural logarithm of eight 𝑥 with respect to 𝑥.

Now, although this may look like a tricky integral to evaluate, it is in fact in a form which we know how to integrate. If we let 𝑓 of 𝑥 be equal to the natural logarithm of eight 𝑥, then we can differentiate 𝑓 of 𝑥 using the fact that the differential of the natural logarithm of 𝑥 is one over 𝑥 in order to find that 𝑓 prime of 𝑥 is equal to one over eight 𝑥. And then, since this is a composite function, we have eight 𝑥 inside the function of the natural logarithm. We mustn’t forget to multiply by the differential of eight 𝑥, which is just eight. This is because of the chain rule. Simplifying, we can obtain that 𝑓 prime of 𝑥 is equal to one over 𝑥.

Now let’s rewrite our integral. If we multiply the numerator and denominator of our fraction by one over 𝑥, then we can rewrite our integral as the integral of negative three over 𝑥 over the natural logarithm of eight 𝑥 with respect to 𝑥. And now we can factor the negative three in the numerator. And once we’ve reached this stage, we notice that this is in a form which we know how to integrate. Since it’s of the form the integral of 𝑎 multiplied by 𝑓 prime of 𝑥 over 𝑓 of 𝑥 d𝑥. Where our 𝑓 of 𝑥 is the natural logarithm of eight 𝑥. And our 𝑓 prime of 𝑥 is one over 𝑥. Therefore, our value of 𝑎 is negative three. Now we know that this integral evaluates to 𝑎 multiplied by the natural logarithm of the absolute value of 𝑓 of 𝑥 plus 𝑐. And we can simply substitute in the values of 𝑎 and 𝑓 of 𝑥 to find our solution. Which is negative three multiplied by the natural logarithm of the absolute value of the natural logarithm of eight 𝑥 plus 𝑐.

In our final example, we’ll be seeing an integration of a trigonometric function where we’re required to multiply our integrand by a fraction of a function over itself.

Determine the indefinite integral of two multiplied by csc of seven 𝑥 with respect to 𝑥.

Now, this is a really tricky function to integrate. It requires us to really know our derivatives of trigonometric functions. Now, since we’re trying to determine the indefinite integral of two csc of seven 𝑥, let’s write down what the derivative of csc of seven 𝑥 is. It is equal to negative seven csc of seven 𝑥 multiplied by cot of seven 𝑥. Now what we’re aiming to do at this stage is to try and get our integrand in the form of 𝑎 multiplied by 𝑓 prime of 𝑥 over 𝑓 of 𝑥. Since we know how to integrate this and in order to try and get our integral in this form. We’re going to try and multiply the integrand by a fraction consisting of a function over itself, which is of course equal to one. The difficult part is trying to find a function 𝑔 of 𝑥. So our integral will end up in the form which we know how to integrate.

Let’s see what happens if we let 𝑔 of 𝑥 be equal to csc of seven 𝑥. We multiply our integrand by csc of seven 𝑥 over csc of seven 𝑥. And we obtain the integral of two csc squared of seven 𝑥 over csc of seven 𝑥 with respect to 𝑥. However, this is clearly not in the form required. However, it does give us a hint since we have a csc squared of seven 𝑥 in the numerator. And we in fact know another trigonometric function which would differentiate to give us a multiple of csc squared of seven 𝑥. And this is cot of seven 𝑥. The derivative of cot of seven 𝑥 is negative seven csc squared of seven 𝑥.

Now it’s at this stage where we might start to be able to spot what our 𝑔 of 𝑥 is. When we multiply by 𝑔 of 𝑥 over 𝑔 of 𝑥 in our integrand, we are always gonna have that factor of csc of seven 𝑥, which is originally in the integrand. If we factor out a factor of csc of seven 𝑥 from our two differentials, then we’re left with a cot of seven 𝑥 multiplied by a constant and a csc of seven 𝑥 multiplied by the same constant. Now, this is very important since the functions which we’re differentiating are csc of seven 𝑥 and cot of seven 𝑥.

And so maybe, for our next 𝑔 of 𝑥, we can try adding these two functions together. But first, let’s make a quick note what the differential of the sum of these two functions is. Using the fact that the differential of a sum of functions is equal to the sum of the differentials of the functions. And by keeping the csc of seven 𝑥 term factored. We have that the differential of csc of seven 𝑥 plus cot of seven 𝑥 is equal to csc of seven 𝑥 multiplied by negative seven cot of seven 𝑥 minus seven csc of seven 𝑥.

So now we can try 𝑔 of 𝑥 being equal to csc of seven 𝑥 plus cot of seven 𝑥. We have the integral of two csc of seven 𝑥 multiplied by csc of seven 𝑥 plus cot of seven 𝑥 over csc of seven 𝑥 plus cot of seven 𝑥 with respect to 𝑥. And we can rewrite this with the two csc of seven 𝑥 in the numerator. Now we can notice that our integral is very nearly in the form which we require. If we let 𝑓 of 𝑥 be equal to 𝑔 of 𝑥, then we can see that the denominator of our integrand is 𝑔 of 𝑥. And if we compare the numerator with the differential of 𝑔 of 𝑥, then we can see that it’s very, very similar. The only two slight differences is that there’s this constant factor of two in the numerator of the integrand and a factor of negative seven in the differential. And so we can come to the conclusion that the numerator of our integrand is equal to two over negative seven times by 𝑔 prime of 𝑥.

We can now more clearly see that our integral is in fact of the form which we require. In our case, 𝑎 is equal to negative two over seven. And 𝑓 of 𝑥 is equal to csc of seven 𝑥 plus cot of seven 𝑥. And so we can apply the formula. Here, we reach our solution, which is that the indefinite integral of two csc of seven 𝑥 with respect to 𝑥 is equal to negative two-sevenths times the natural logarithm of the absolute value of csc of seven 𝑥 plus cot of seven 𝑥 plus 𝑐.

In this last example, we saw how we can sometimes find the integral of a difficult function using this method by multiplying by 𝑔 of 𝑥 over 𝑔 of 𝑥 for some function 𝑔 of 𝑥. And the most difficult part of this method is working out what 𝑔 of 𝑥 needs to be. Knowing the differentials of many different types of functions is certainly a very useful tool in doing this.

We have now covered a variety of examples in this video. Let’s recap some of the key points. Key points. The indefinite integral of 𝑎 multiplied by 𝑓 prime of 𝑥 over 𝑓 of 𝑥 with respect to 𝑥 is equal to 𝑎 multiplied by the natural logarithm of the absolute value of 𝑓 of 𝑥 plus 𝑐, where 𝑎 is a real number. This method can be used to find the integrals of many different types of functions, including trigonometric ones such as cot 𝑥, tan 𝑥, csc 𝑥, and so on. Sometimes it is not immediately obvious that we can use this method to integrate. But multiplying by 𝑔 of 𝑥 over 𝑔 of 𝑥 for some 𝑔 of 𝑥 may enable us to use it.