### Video Transcript

Determine the integral of the natural logarithm of three π₯ minus five with respect to π₯.

Weβre asked to determine an integral where our integrand is the composition of two functions. And thereβs a few different ways we could approach this problem. For example, we could try directly using integration by parts. And this would work. However, thereβs a simpler method. Since our integrand is the composition of two functions, weβll try integrating this by using substitution. Weβll substitute π’ is equal to three π₯ minus five. If π’ is equal to three π₯ minus five, we differentiate both sides with respect to π₯. We get ππ’ by dπ₯ is equal to the coefficient of π₯, which is three.

And remember, we know that ππ’ by ππ₯ is not a fraction. However, when weβre using integration by substitution, we can treat it a little bit like a fraction. This gives us the equivalent statement in terms of differentials ππ’ is equal to three ππ₯.

Weβre now ready to use our substitution. First, we have three π₯ minus five is equal to π’. And by dividing through by three, we see that one-third ππ’ is equal to ππ₯. This gives us the integral of the natural logarithm of π’ times one-third ππ’. And now, we can rearrange this equation and take the factor of one-third outside of our integral to give us one-third times the integral of the natural logarithm of π’ with respect to π’. And while we might not know an antiderivative for the natural logarithm of π’, we know we can find this by using integration by parts.

Before we use integration by parts, remember, this is an integral in terms of π’. So, we canβt call one of our functions in integration by parts π’. So instead, weβll call this π€. Integration by parts tells us if π€ and π£ prime are functions of π’, then the integral of π€π£ prime with respect to π’ is equal to π€ times π£ minus the integral of π£ times π€ prime with respect to π’.

And we might be worried about a problem here. Our integrand is not the product of two functions. However, we can fix this by writing our integrand as the natural logarithm of π’ all multiplied by one. We now need to remember when weβre using integration by parts, we want to choose π€ so that when we differentiate it, we get something more simple. And weβve heard a few different ways of choosing this function.

For example, we could use the LIATE method. The L in LIATE tells us to choose the logarithmic function. So, we would choose π€ is the natural logarithm of π’. However, itβs not necessary to use this in this case since we need to integrate our function π£ prime. And we already said we donβt know how to integrate the natural logarithm of π’ directly. So, we need to pick π£ prime equal to one. So, weβll set π€ equal to the natural logarithm of π’ and π£ prime equal to one.

To use integration by parts, we need expressions for π€ prime and π£. Letβs start with π€ prime. π€ prime will be the derivative of the natural logarithm of π’ with respect to π’. We know this is just equal to one over π’. Next, to find π£, we know the integral of one with respect to π’ will be equal to π’ plus a constant of integration. We just need any antiderivative. So, weβll just pick π’. We can now substitute our expressions for π€, π£, π€ prime, and π£ prime into our expression for integration by parts.

This gives us one-third multiplied by the natural logarithm of π’ times π’ minus the integral of π’ multiplied by one over π’ with respect to π’. And now, we can start simplifying. First, in our integrand, we have π’ divided by π’. So, the integrand simplifies to give us one. So, with some rearranging, we get one-third times π’ multiplied by the natural logarithm of π’ minus the integral of one with respect to π’. And we know negative one multiplied by the integral of one with respect to π’ would be negative π’ plus a constant of integration.

However, to make things simpler, weβll just write negative π’ and then add a constant of integration at the end of our expression. This gives us one-third times π’ multiplied by the natural logarithm of π’ minus π’ plus our constant of integration πΆ. We can still simplify our answer even more. Weβll take out the constant factor of π’. This gives us one-third times π’ multiplied by the natural logarithm of π’ minus one plus our constant of integration πΆ.

Finally, we need to remember that this is an expression for an integral in terms of π₯. So, we should give our answer in terms of π₯. To do this, weβll use our substitution π’ is equal to three π₯ minus five. So, by substituting π’ is equal to three π₯ minus five, we get one-third times three π₯ minus five multiplied by the natural logarithm of three π₯ minus five minus one plus πΆ. Therefore, by using integration by substitution first and then using integration by parts, we were able to show the integral of the natural logarithm of three π₯ minus five with respect to π₯ is equal to one-third times three π₯ minus five multiplied by the natural logarithm of three π₯ minus five minus one plus πΆ.