# Video: APCALC02AB-P1A-Q28-805131059291

Calculate β«_(1) ^(π) ((ln π₯)β΄/π₯) dπ₯.

02:06

### Video Transcript

Calculate the integral evaluated between one and π of the natural log of π₯ to the fourth power divided by π₯ with respect to π₯.

Weβre going to begin by rewriting our integral slightly. We can say that the natural log of π₯ to the fourth power divided by π₯ is the same as the natural log of π₯ of the fourth power times one over π₯. And then, we notice that the derivative of one part of our integral is equal to another part. The derivative of the natural log of π₯ with respect to π₯ is equal to one over π₯. This tells us we can use integration by substitution to solve this problem. Weβre going to let π’ be equal to the natural log of π₯. And weβve then seen that the derivative of π’ with respect to π₯ is equal to one over π₯.

Now, dπ’ by dπ₯ is not a fraction. But for the purposes of integration by substitution, we can say that dπ’ by dπ₯ equals one over π₯ is equivalent to dπ’ equals one over π₯ dπ₯. And we can now see we can replace one over π₯ dπ₯ with dπ’. And we can replace the natural log of π₯ with π’. And weβll be integrating π’ to the fourth power with respect to π’. But what about our limits?

Our limits are currently in terms of π₯. So weβll use the fact that we let π’ be equal to the natural log of π₯. And when π₯ is equal to π, we know that π’ is going to be equal to the natural log of π, which is of course one. This also means that when π₯ is equal to one, π’ is equal to the natural log of one which is of course zero. And so, we can now evaluate the integral between zero and one of π’ to the fourth power with respect to π’. To integrate π’ to the fourth power, we add one to the power and divide by that new value.

So the integral of π’ to the fourth power is π’ to the fifth power divided by five. And weβre going to evaluate this between the limits of zero and one. Thatβs one to the fifth power divided by five minus zero to the fifth power divided by five which is equal to one-fifth. And weβre done.

We used π’ is equal to the natural log of π₯ to change our limits. So we can see that the integral evaluated between one and π of the natural log of π₯ to the fourth power divided by π₯ with respect to π₯ is equal to one-fifth.