### Video Transcript

Which of the following is equivalent to the integral from one to π cubed of three plus five times the natural logarithm of π₯ all raised to the power of four divided by π₯ with respect to π₯. Listed answer a) one over five multiplied by the integral from one to eight of π’ to the fourth power with respect to π’. b) One over five multiplied by the integral from three to 18 of π’ to the fourth power with respect to π’. c) The integral from three to 18 of π’ to the fourth power with respect to π’. Or d) The integral from three to eight of π’ to the fourth power with respect to π’.

Notice that the original integral uses the variable π₯. And the options for the answer use the variable π’. To introduce a new variable into the original integral, we will use the method of integration by substitution. We see that each of the listed options contain the fourth power of π’. So letβs introduce the function π’ of π₯ to be equal to three plus five times the natural logarithm of π₯. Then, π’ to the power of four is equal to three plus five times the natural logarithm of π₯ all raised to the power of four, which is a term we can recognise in the original integral.

Before we substitute the term three plus five times the natural logarithm of π₯ in the original integral with π’, letβs find dπ’ by dπ₯, the derivative of π’ with respect to π₯. In order to find dπ’ by dπ₯, note that π’ of π₯ is a sum of the constant function three and the function five times the natural logarithm of π₯. So using the fact that the derivative of a sum of functions is equal to the sum of their derivatives, in order to differentiate π’ with respect to π₯, we will differentiate each summand in π’ separately with respect to π₯ and add the derivatives together.

We have that the derivative of the constant function three with respect to π₯ is the zero function. Using the fact that we can pull constants out whilst differentiating and the fact that the derivative of the natural logarithm of π₯ with respect to π₯ is one over π₯, we have that the derivative of the function five times the natural logarithm of π₯ is five times one over π₯. Simplifying, we obtain that dπ’ by dπ₯ equals five over π₯.

The next step is to treat dπ’ by dπ₯ like a fraction although it isnβt one and make dπ₯ the subject. Doing so, we obtain that dπ₯ is equal to π₯ over five dπ’. Substituting the expressions for three plus five times the natural logarithm of π₯ and dπ₯ in terms of π’ in the original integral, we obtain that, without the limits, the original integral is equal to the integral of π’ to the fourth power over π₯ multiplied by π₯ over five with respect to π’.

We can remove the variable π₯ from the integrand by cancelling it out on the numerator and denominator and pull the constant one over five outside the integral. Now, letβs include the new limits of integration, which are formed by applying the function π’ to the original limits of integration. The lower limit is π’ of one. And the upper limit is π’ of π cubed. π’ of one equals three as the natural logarithm of one equals zero. π’ of π cubed equals three plus five times the natural logarithm of π cubed.

Using the fact that the natural logarithm of π to the power of π equals π times the natural logarithm of π, we can rewrite the natural logarithm of π cubed as three times the natural logarithm of π. Since the natural logarithmic function and the exponential function π are inverses of each other, we have that the natural logarithm of π is equal to one. Therefore, π’ of π cubed is equal to three plus five times three, which is three plus 15, which is equal to 18. Therefore, our original integral is equivalent to one over five multiplied by the integral from three to 18 of π’ to the fourth power with respect to π’.

We see that this is the same as option b). This is the final answer to the question. However, going one step further, letβs quickly take a moment to see how we could evaluate this integral. We can integrate π’ to the fourth power with respect to π’ by increasing the power by one and dividing by the new power to obtain π’ to the fifth power over five. Evaluating this between three and 18 and then dividing by five, we obtain 75573. So we can evaluate the integral in this way. However, this is of course only an additional exercise.