Video: Finding the Integration of a Function Involving a Logarithmic Function Using Integration by Substitution

Determine ∫ (βˆ’11/6π‘₯) βˆ›(ln π‘₯) dπ‘₯.

01:46

Video Transcript

Determine the integral of negative 11 over six π‘₯ times the cube root of the natural log of π‘₯ dπ‘₯.

In order to evaluate this integral, we need to spot that the derivative of the natural log of π‘₯ is one over π‘₯ and that a part of our function is a scalar multiple of one over π‘₯. This tells us we can use integration by substitution to evaluate our integral. We let 𝑒 be equal to the natural log of π‘₯. And we’ve seen that d𝑒 by dπ‘₯ is, therefore, one over π‘₯.

Remember, d𝑒 by dπ‘₯ 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 d𝑒 is equal to one over π‘₯dπ‘₯. 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 π‘₯ dπ‘₯ with d𝑒. And we can replace the natural log of π‘₯ with 𝑒.

To make this easy to integrate, we recall that the cube root of 𝑒 is the same as 𝑒 to the power of one-third. And we know that the antiderivative of 𝑒 to the power of one-third is 𝑒 to the power of four-thirds divided by four-thirds, or three-quarters 𝑒 to the power of four-thirds. We distribute our parentheses, and we see that our integral is equal to negative eleven-eighths times 𝑒 to the power of four-thirds plus capital 𝐢.

And I’ve changed it from a lowercase 𝑐 to a capital 𝐢, 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 π‘₯. So, we replace 𝑒 with the natural log of π‘₯. And we see that our answer is negative eleven-eighths times the natural log of π‘₯ to the power of four-thirds plus 𝐢.

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