WEBVTT
00:00:00.790 --> 00:00:09.130
Determine the indefinite integral of π₯ squared plus seven over π₯ cubed plus 21π₯ minus five with respect to π₯.
00:00:11.190 --> 00:00:15.540
In order to solve this problem, we will be using integration by substitution.
00:00:16.560 --> 00:00:21.780
The way that we can spot this is if we take our denominator and call it π of π₯.
00:00:22.680 --> 00:00:27.360
So we have π of π₯ is equal to π₯ cubed plus 21π₯ minus five.
00:00:28.450 --> 00:00:31.510
Now, we differentiate this to find π prime of π₯.
00:00:32.570 --> 00:00:35.660
Differentiating the π₯ cubed, we get three π₯ squared.
00:00:36.470 --> 00:00:42.570
Differentiating 21π₯, we get 21 and minus five differentiates to give zero.
00:00:43.700 --> 00:00:55.230
So π prime of π₯ is equal to three π₯ squared plus 21, which we can also write as three times by π₯ squared plus seven since we just factor out the three.
00:00:56.120 --> 00:01:05.230
And at this point, we notice that we have some constant which is three multiplied by the numerator of our fraction in the integral.
00:01:06.420 --> 00:01:15.000
Now, we can rearrange this to write that π₯ squared plus seven is equal to one-third times π prime of π₯.
00:01:16.060 --> 00:01:24.810
So we can write our integral as one-third timesed by π prime of π₯ over π of π₯ with respect to π₯.
00:01:25.770 --> 00:01:29.870
And since that one-third is just a constant, we can factor out of the integral.
00:01:30.750 --> 00:01:32.480
And so that leaves us with this.
00:01:34.390 --> 00:01:49.350
Now, our integral is of the form: the integral of πΉ of π of π₯ times π prime of π₯ with respect to π₯, where in our case the πΉ is πΉ of π of π₯ is equal to one over π of π₯.
00:01:50.490 --> 00:01:55.370
So now, our integral is of this form, we can use what is called a π’ substitution.
00:01:56.520 --> 00:02:00.690
And in this substitution, weβre going to let π’ equal π of π₯.
00:02:01.550 --> 00:02:05.320
Now, weβre going to need to find ππ’ in terms of ππ₯.
00:02:06.190 --> 00:02:12.440
We can use the fact that ππ’ is equal to ππ’ by ππ₯ times ππ₯.
00:02:13.410 --> 00:02:19.520
Now, since π’ is equal to π of π₯, ππ’ ππ₯ is therefore equal to π prime of π₯.
00:02:20.530 --> 00:02:26.290
And so we get ππ’ is equal to π prime of π₯ times ππ₯.
00:02:27.480 --> 00:02:41.790
We can rewrite our integral one-third lots of the integral times π prime of π₯ over π of π₯ ππ₯ as one-third lots of the integral of one over π of π₯ times π prime of π₯ with respect to π₯.
00:02:42.550 --> 00:02:50.140
And now, we notice that we have π prime of π₯ times π of π₯ which is the same as ππ’ of π₯.
00:02:51.220 --> 00:02:59.300
So we are ready to make the substitution, remembering that π’ equals π of π₯ and ππ’ is equal to π prime times π of π₯.
00:03:00.230 --> 00:03:06.240
And so we get that this is equal to one-third times the integral of one over π’ ππ’.
00:03:07.310 --> 00:03:12.560
Now, all we need to do is calculate one-third times the integral of one over π’ ππ’.
00:03:13.450 --> 00:03:18.900
And we get that is equal to one-third times the natural logarithm of the absolute value of π’.
00:03:19.720 --> 00:03:24.540
Now, since this was an indefinite integral, we mustnβt forget to add on the constant of integration.
00:03:24.900 --> 00:03:27.060
So we get plus π on the end.
00:03:28.570 --> 00:03:31.340
Next, we can do the inverse of our substitution.
00:03:32.180 --> 00:03:34.400
So we have that π’ is equal to π of π₯.
00:03:34.530 --> 00:03:42.630
So we can substitute this back in, giving us one-third times the natural logarithm of the absolute value of π of π₯ plus π.
00:03:43.780 --> 00:03:50.720
And for our final step, we can substitute back in that π of π₯ is equal to π₯ cubed plus 21π₯ minus five.
00:03:51.640 --> 00:04:08.340
We get the final answer that the indefinite integral of π₯ squared plus seven over π₯ cubed plus 21π₯ minus five with respect to π₯ is equal to one-third times the natural logarithm of the absolute value of π₯ cubed plus 21π₯ minus five plus π.