# Basic Integration: Exercise 2

by Batool Akmal

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00:01 Let's have a look at our second integral.

00:04 You can see here that this looks like a function of a function equation.

00:09 So let's see how we deal with this.

00:11 Remember, we've done a very similar example in the previous lecture.

00:16 So let's attempt to integrate this.

00:18 It's still an indefinite integral, so there are no limits.

00:21 That means we will put a plus C at the end and it's to the power of three dx.

00:27 So remember what we said.

00:28 If you're integrating something like this, we're trying to do the exact opposite of the chain rule when we differentiate.

00:34 So, this is the outside function, this is the inside function.

00:40 To start off, we're going to integrate the outside function without changing anything on the inside.

00:46 So to integrate, you add one to the power, divide by new power.

00:50 And then rather than multiplying with the differential of the inside, the differential of the inside in this case is 10.

00:57 We are going to divide it by the differential of the inside, and obviously those two values at the bottom multiply.

01:03 So we are adding one to the power, dividing by new power, dividing by the differential of the inside.

01:10 Put plus C there or later.

01:12 So we've got 10x minus 5 to the power of 4 all over 40 plus C.

The lecture Basic Integration: Exercise 2 by Batool Akmal is from the course Basic Integration.

### Included Quiz Questions

1. [(3x + 1)³ / 9] + c
2. [(3x + 1)³ / 30] + c
3. [(3x + 1)³ / 3] + c
4. [(3x + 1)³ / 9]
5. [(3x + 1)³ / 90] + c
1. [(10x -4)²³ / 230] + c
2. [(10x - 4)²³ / 23] + c
3. [(10x - 4)²³ / 230]
4. [(10x + 4)²³ / 230] + c
5. [(10x - 4)²² / 230] + c

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