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

Determine βˆ«βˆ’(7 ln 3π‘₯)/(5π‘₯ ln √(3π‘₯)) dπ‘₯.

02:55

Video Transcript

Determine the integral of negative seven times the natural logarithm of three π‘₯ divided by five π‘₯ multiplied by the natural logarithm of the square root of three π‘₯ with respect to π‘₯.

We can see this question is asking us to evaluate the integral of a quotient of two functions. This is a very complicated integral which we can’t do directly. So, our first instinct might be to try using integration by substitution or integration by parts, or maybe some other method of integration. However, it’s often very useful to check to see if we can simplify our integrand into something we can integrate.

In this instance, we can see we have the natural logarithm of three π‘₯ in our numerator and the natural logarithm of the square root of three π‘₯ in our denominator. So, one thing we could try using is our laws of logarithms to simplify our integrand. In particular, let’s focus on the natural logarithm of the square root of three π‘₯. First, by using our laws of exponents, we can rewrite the square root of three π‘₯ as three π‘₯ all raised to the power of one-half. But then, since the natural logarithm is just the logarithm base 𝑒, we can rewrite this by using our power rule for logarithms.

We get the natural logarithm of three π‘₯ to the power of one-half is just equal to one-half times the natural logarithm of three π‘₯. And if we have this in the denominator of our integrand, we can see both the denominator and the numerator of our integrand share a factor of the natural logarithm of three π‘₯. So by doing this, we can cancel these and we’ll simplify our integrand. So let’s use this to rewrite our integral.

First, we’ll replace the natural logarithm of the square root of three π‘₯ in the denominator of our integrand with one-half times the natural logarithm of three π‘₯. Next, we’ll cancel the shared factor of the natural logarithm of three π‘₯ in the numerator and the denominator of our integrand. Finally, instead of dividing by one-half, we’ll multiply by the reciprocal of one-half. This is the same as multiplying by two. So, we have the integral of negative 14 divided by five π‘₯ with respect to π‘₯.

And now, we can see that this is an integral in a standard form. This is the integral of negative 14 divided by five multiplied by the reciprocal function with respect to π‘₯. And we know the integral of π‘Ž over π‘₯ with respect to π‘₯ is equal to π‘Ž times the natural logarithm of the absolute value of π‘₯ plus the constant of integration 𝐢. In our case, the value of π‘Ž is negative 14 over five. So, if we apply this, we get our final answer, negative 14 over five times the natural logarithm of the absolute value of π‘₯ plus 𝐢.

Therefore, by simplifying our integrand by using our logarithm laws, we’ve shown the integral of negative seven times the natural logarithm of three π‘₯ divided by five π‘₯. Times the natural logarithm of the square root of three π‘₯ with respect to π‘₯ is equal to negative 14 over five times the natural logarithm of the absolute value of π‘₯ plus 𝐢.

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