Video: Finding the Integration of a Function Involving Using the Laws of Logarithms

Determine βˆ«βˆ’(6 ln π‘₯Β²)/(5π‘₯ ln π‘₯Β³) dπ‘₯.

02:47

Video Transcript

Determine the integral of negative six times the natural logarithm of π‘₯ squared all divided by five π‘₯ times the natural logarithm of π‘₯ cubed with respect to π‘₯.

This question is asking us to evaluate the integral of the quotient of two functions. We see we can’t evaluate this integral directly. So, we’re going to need to use one of our tools. We might be tempted to try using a 𝑒 substitution or using integration by parts. However, the first thing we should always check is, can we simplify our integrand? We see both the numerator and denominator of our integrand contain a logarithm. So, we’ll try rewriting this by using our log laws.

We can rewrite both of these logarithmic terms by using the power rule for logarithms, which tells us the logarithm of π‘Ž to the 𝑛th power is equal to 𝑛 times the logarithm of π‘Ž. Since the natural logarithm is a logarithm of base 𝑒, we can use this rule on our natural logarithm function. Using this, we get the natural logarithm of π‘₯ squared is equal to two times the natural logarithm of π‘₯ and the natural logarithm of π‘₯ cubed is equal to three times the natural logarithm of π‘₯.

Let’s now use both of these to rewrite our integral. We get the integral of negative six times two times the natural logarithm of π‘₯ divided by five π‘₯ multiplied by three times the natural logarithm of π‘₯ with respect to π‘₯. And we can now simplify this expression. We have a shared factor of the natural logarithm of π‘₯ in our numerator and our denominator. So, we can cancel these out. We also have a shared factor of three in our numerator and our denominator.

We can then simplify our numerator to get the integral of negative four divided by five π‘₯ with respect to π‘₯. And we could just directly evaluate this integral. However, we’ll do this in full. We’ll take out the constant factor of negative four over five outside of our integral. So, this leaves us to evaluate the integral of the reciprocal function with respect to π‘₯. And we know this. It’s equal to the natural logarithm of the absolute value of π‘₯ plus the constant of integration 𝐢.

Applying this to our integral, we get negative four-fifths times the natural logarithm of the absolute value of π‘₯ plus the constant of integration 𝐢 one. Next, we’ll distribute negative four-fifths over our parentheses. We get negative four-fifths times the natural logarithm of the absolute value of π‘₯ minus four 𝐢 one over five. And the last thing we’ll do is simplify our constant. Since 𝐢 one is a constant, negative four 𝐢 one divided by five is also a constant. So, we’ll rewrite this entire constant as 𝐢.

And this gives us our final answer. The integral of negative six times the natural logarithm of π‘₯ squared divided by five π‘₯ times the natural logarithm of π‘₯ cubed with respect to π‘₯. Is equal to negative four-fifths times the natural logarithm of the absolute value of π‘₯ plus 𝐢.

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