Question Video: Finding the Integration of a Logarithmic Function Mathematics • Higher Education

Determine ∫ ln(3π‘₯ βˆ’ 5) dπ‘₯.

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Video Transcript

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

We’re asked to determine an integral where our integrand is the composition of two functions. And there’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, there’s a simpler method. Since our integrand is the composition of two functions, we’ll try integrating this by using substitution. We’ll substitute 𝑒 is equal to three π‘₯ minus five. If 𝑒 is equal to three π‘₯ minus five, we differentiate both sides with respect to π‘₯. We get 𝑑𝑒 by dπ‘₯ is equal to the coefficient of π‘₯, which is three.

And remember, we know that 𝑑𝑒 by 𝑑π‘₯ is not a fraction. However, when we’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 𝑑𝑒 is equal to three 𝑑π‘₯.

We’re now ready to use our substitution. First, we have three π‘₯ minus five is equal to 𝑒. And by dividing through by three, we see that one-third 𝑑𝑒 is equal to 𝑑π‘₯. This gives us the integral of the natural logarithm of 𝑒 times one-third 𝑑𝑒. 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 𝑒 with respect to 𝑒. And while we might not know an antiderivative for the natural logarithm of 𝑒, 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 𝑒. So, we can’t call one of our functions in integration by parts 𝑒. So instead, we’ll call this 𝑀. Integration by parts tells us if 𝑀 and 𝑣 prime are functions of 𝑒, then the integral of 𝑀𝑣 prime with respect to 𝑒 is equal to 𝑀 times 𝑣 minus the integral of 𝑣 times 𝑀 prime with respect to 𝑒.

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 𝑒 all multiplied by one. We now need to remember when we’re using integration by parts, we want to choose 𝑀 so that when we differentiate it, we get something more simple. And we’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 𝑀 is the natural logarithm of 𝑒. However, it’s not necessary to use this in this case since we need to integrate our function 𝑣 prime. And we already said we don’t know how to integrate the natural logarithm of 𝑒 directly. So, we need to pick 𝑣 prime equal to one. So, we’ll set 𝑀 equal to the natural logarithm of 𝑒 and 𝑣 prime equal to one.

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

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

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

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, we’ll use our substitution 𝑒 is equal to three π‘₯ minus five. So, by substituting 𝑒 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 𝐢. 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 𝐢.

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