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

Determine β« (β11/6π₯) β(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 π’ be equal to the natural log of π₯. And weβve seen that dπ’ by dπ₯ is, therefore, one over π₯.

Remember, dπ’ 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 dπ’ is equal to one over π₯dπ₯. Letβ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 dπ’. And we can replace the natural log of π₯ with π’.

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

And Iβve changed it from a lowercase π to a capital πΆ, as weβ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 π’ 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 πΆ.

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