Question Video: Finding the Integration of a Logarithmic Function | Nagwa Question Video: Finding the Integration of a Logarithmic Function | Nagwa

# Question Video: Finding the Integration of a Logarithmic Function Mathematics • Third Year of Secondary School

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Determine β« ln(3π₯ β 5) dπ₯.

04:42

### Video Transcript

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

Weβre asked to determine an integral where our integrand is the composition of two functions. And thereβs a few different ways we could approach this problem. For example, we could try directly using integration by parts. And this would work. However, thereβs a simpler method. Since our integrand is the composition of two functions, weβll try integrating this by using substitution. Weβll 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 ππ’ by dπ₯ is equal to the coefficient of π₯, which is three.

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

Weβre 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 ππ’ is equal to ππ₯. This gives us the integral of the natural logarithm of π’ times one-third ππ’. 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 canβt call one of our functions in integration by parts π’. So instead, weβll 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 weβre using integration by parts, we want to choose π€ so that when we differentiate it, we get something more simple. And weβve 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, itβs not necessary to use this in this case since we need to integrate our function π£ prime. And we already said we donβt know how to integrate the natural logarithm of π’ directly. So, we need to pick π£ prime equal to one. So, weβll 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£. Letβs 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, weβll 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, weβll 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 πΆ. We can still simplify our answer even more. Weβll 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 πΆ.

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, weβll 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 πΆ. 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 πΆ.

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