### Video Transcript

Determine the integral of three divided by seven π₯ times the natural logarithm of π₯ with respect to π₯.

In this question, weβre asked to evaluate the indefinite integral of the quotient of two functions. And we can recall evaluating the integral of the quotient of two functions is very difficult. So letβs try and use one of our integral rules to help us evaluate this integral. We can recall the following result involving the integral of the quotient of two functions. If we want to evaluate the integral of π prime of π₯ divided by π of π₯ with respect to π₯, then we know this is equal to the natural logarithm of the absolute value of π of π₯ plus a constant of integration πΆ.

To apply this result to our integral, we need to check itβs of the desired form. So letβs set π of π₯ to be the function in the denominator. Thatβs seven π₯ times the natural logarithm of π₯. We now need to find π prime of π₯. Thatβs the derivative of π of π₯ with respect to π₯. Since π of π₯ is the product of two functions, we can do this by using the product rule for differentiation.

And we can recall the product rule for differentiation tells us we can just differentiate each factor separately and find the sum of these expressions. π prime of π₯ is equal to the derivative of seven π₯ with respect to π₯ times the natural logarithm of π₯ plus seven π₯ multiplied by the derivative of the natural logarithm of π₯ with respect to π₯. We know the derivative of seven π₯ with respect to π₯ is the coefficient of π₯, which is seven. And we can recall the derivative of the natural logarithm of π₯ with respect to π₯ is the reciprocal function one over π₯. So in this case, π prime of π₯ is seven times the natural logarithm of π₯ plus seven π₯ divided by π₯.

And this is where we can see a problem. Even if we cancel the shared factor of π₯ in this expression, we get seven times the natural logarithm of π₯ plus seven. This is not the expression in the numerator, so we might conclude that we canβt use this rule to help us evaluate this integral. However, this is not the case. We just need to be more careful with our choice of our function π of π₯. To do this, we need to notice if we chose π of π₯ to be the natural logarithm of π₯ with respect to π₯, then we know the derivative of this function is one divided by π₯. And we can see this also appears in our integrand.

So we can apply this rule by first rewriting our integrand to have a denominator of the natural logarithm of π₯. Weβll do this by clearing some space and then taking three divided by seven π₯ into the numerator of our expression. This gives us the integral of three over seven π₯ all divided by the natural logarithm of π₯ with respect to π₯. Now, if we set our function π of π₯ to be the denominator of this integrand, thatβs the natural logarithm of π₯, then π prime of π₯ is equal to one divided by π₯. And this is almost exactly equal to the expression in our numerator. However, we have a constant factor of three-sevenths. So letβs take this outside of our integral. This gives us three-sevenths times the integral of one over π₯ divided by the natural logarithm of π₯ with respect to π₯.

And now this is exactly in the form of our integral rule with π of π₯ equal to the natural logarithm of π₯. And now we can just apply our integral rule to evaluate this integral. We get three-sevenths times the natural logarithm of the absolute value of the natural logarithm of π₯ plus a constant of integration πΆ, which is our final answer.

Therefore, we were able to show the integral of three divided by seven π₯ times the natural logarithm of π₯ with respect to π₯ is three-sevenths times the natural logarithm of the absolute value of the natural logarithm of π₯ plus a constant of integration πΆ.