Determine the integral of negative 11 over six 𝑥 times the cube root of the natural log of 𝑥 d𝑥.
In order to evaluate this integral, we need to spot that the derivative of the natural log of 𝑥 is one over 𝑥 and that a part of our function is a scalar multiple of one over 𝑥. This tells us we can use integration by substitution to evaluate our integral. We let 𝑢 be equal to the natural log of 𝑥. And we’ve seen that d𝑢 by d𝑥 is, therefore, one over 𝑥.
Remember, d𝑢 by d𝑥 is not a fraction but we treat it a little like one when performing integration by substitution. And we see that this is equivalent to saying d𝑢 is equal to one over 𝑥d𝑥. Let’s substitute what we now have into our integral. If we take out a factor of negative eleven-sixths, we see that we can replace one over 𝑥 d𝑥 with d𝑢. And we can replace the natural log of 𝑥 with 𝑢.
To make this easy to integrate, we recall that the cube root of 𝑢 is the same as 𝑢 to the power of one-third. And we know that the antiderivative of 𝑢 to the power of one-third is 𝑢 to the power of four-thirds divided by four-thirds, or three-quarters 𝑢 to the power of four-thirds. We distribute our parentheses, and we see that our integral is equal to negative eleven-eighths times 𝑢 to the power of four-thirds plus capital 𝐶.
And I’ve changed it from a lowercase 𝑐 to a capital 𝐶, as we’ve multiplied our original constant of integration by negative eleven-sixths, thereby changing the number. It is, of course, important to remember that our original integral was in terms of 𝑥. So, we replace 𝑢 with the natural log of 𝑥. And we see that our answer is negative eleven-eighths times the natural log of 𝑥 to the power of four-thirds plus 𝐶.