iGCSE (2021 Edition)

# 12.02 The reverse chain rule

Lesson

We saw in the previous lesson how to integrate (find the anti-derivative of) functions made up of multiple terms. In this lesson, just as we did with differentiation, we will expand our capabilities to include different types of functions. Remember, the structure of the function being differentiated or integrated determines the method used.

## Integration of $(ax+b)^n$(ax+b)n

To differentiate a linear function raised to a power we would use the chain rule and follow the following steps:

1. Reduce the original index by one
2. Multiply by the original index
3. Multiply by the derivative of the function inside the brackets

We have established that integration is the reverse of differentiation and hence reversing the steps above:

1. Increase the index by one
2. Divide by the new index
3. Divide by the derivative of the function inside the brackets

To find $\int\ \left(2x-3\right)^5\ dx$ (2x3)5 dx we will consider the derivative of the function with an index one higher than our expression to be integrated, that is, . We differentiate this function with respect to $x$x as follows:

 $\frac{d}{dx}\left(2x-3\right)^6$ddx​(2x−3)6 $=$= $6\left(2x-3\right)^5\times2$6(2x−3)5×2 $=$= $12\left(2x-3\right)^5$12(2x−3)5

Hence we could say that $12\left(2x-3\right)^5$12(2x3)5 is the derivative of $\left(2x-3\right)^6$(2x3)6, and $\left(2x-3\right)^6$(2x3)6 is an anti-derivative of $12\left(2x-3\right)^5$12(2x3)5.

We can use this result to explore the integral of $\left(2x-3\right)^5$(2x3)5:

 $\int\ \left(2x-3\right)^5\ dx$∫ (2x−3)5 dx $=$= $\frac{1}{12}\int\ 12\left(2x-3\right)^5\ dx$112​∫ 12(2x−3)5 dx

We have multiplied the function to be integrated by one more than the index and the derivative of the term inside the brackets, in this case $12$12. We balance this out by multiplying the whole integral by an equivalent factor to "cancel out" this factor, in this case $\frac{1}{12}$112.

Using our previous result of $\frac{d}{dx}(2x-3)^6=12(2x-3)^5$ddx(2x3)6=12(2x3)5 and reversing the differentiation:

 $\frac{1}{12}\int\ 12(2x-3)^5\ dx$112​∫ 12(2x−3)5 dx $=$= $\frac{1}{12}(2x-3)^6+C$112​(2x−3)6+C

Note that the factor of $\frac{1}{12}$112 is included in our final result. Let's look at the reason for this by differentiating our integral to see if we end up with our original expression $(2x-3)^5$(2x3)5:

 $\frac{d}{dx}\frac{(2x-3)^6}{12}+C$ddx​(2x−3)612​+C $=$= $\frac{1}{12}\times6\times2\times(2x-3)^5$112​×6×2×(2x−3)5 $=$= $(2x-3)^5$(2x−3)5

Hence, we can see that differentiating our integrated expression gives our original expression. Pay careful attention, when integrating, to the step of dividing by the new index and derivative of the expression in the brackets. We can see that this factor becomes an adjustment factor of sorts to ensure that integral and its associated derivative are compatible.

The following rule can be used to integrate linear functions raised to a power:

Integration of $(ax+b)^n$(ax+b)n

$\int\ (ax+b)^ndx=\frac{(ax+b)^{n+1}}{a(n+1)}+C$ (ax+b)ndx=(ax+b)n+1a(n+1)+C for $n\ne-1$n1

#### Worked example

##### Example 1

Find $\int\left(2x+1\right)^3\ dx$(2x+1)3 dx.

Think: This is a linear function raised to a power so we can use the rule above to integrate this expression.

Do: We increase the power by one, divide by the new power and the derivative of the linear function inside the brackets.

 $\int\left(2x+1\right)^3\ dx$∫(2x+1)3 dx $=$= $\frac{\left(2x+1\right)^4}{4\times2}+C$(2x+1)44×2​+C $=$= $\frac{\left(2x+1\right)^4}{8}+C$(2x+1)48​+C
##### Example 2

Find $\int\frac{2}{\left(x+1\right)^4}\ dx$2(x+1)4 dx.

Think: There is a linear function raised to a power in the denominator of this expression so we can't use the rule developed above yet. However, if we manipulate this expression using negative indices we can get it into a form that we can integrate.

Do: We manipulate the expression into an appropriate form using negative indices and then integrate using the reverse chain rule:

 $\int\frac{2}{\left(x+1\right)^4}\ dx$∫2(x+1)4​ dx $=$= $\int\ 2\left(x+1\right)^{-4}\ dx$∫ 2(x+1)−4 dx $=$= $\frac{2\left(x+1\right)^{-3}}{-3\times1}+C$2(x+1)−3−3×1​+C $=$= $\frac{2}{-3\left(x+1\right)^3}+C$2−3(x+1)3​+C

It is normal practice to rewrite the expression without negative indices.

Note that the coefficient of $2$2 is treated in the same way as the process of differentiation. It is treated as a constant multiplier throughout the integration process and can actually be taken outside the integral. That is, $2\ \int\frac{1}{\left(x+1\right)^4}\ dx$2 1(x+1)4 dx would give the same result as was obtained above.

##### Example 3

Find $\int\ \left(x^2-1\right)^2\ dx$ (x21)2 dx.

Think: We are asked to integrate the square of a non-linear expression and can't use the rule above in this case. However, if we expand this expression we can get it into a form that we can integrate.

Do: We expand the expression and then integrate each term individually:

 $\int\ \left(x^2-1\right)^2\ dx$∫ (x2−1)2 dx $=$= $\int\ x^4-2x^2+1\ dx$∫ x4−2x2+1 dx $=$= $\frac{x^5}{5}-\frac{2x^3}{3}+x+C$x55​−2x33​+x+C

#### Practice question

##### Question 1

Find $\int\left(2x-5\right)^{\frac{1}{2}}dx$(2x5)12dx, with $C$C as the constant of integration.

##### Question 2

Find $\int\left(\sqrt{x-2}+\sqrt{x+3}\right)dx$(x2+x+3)dx, with $C$C as the constant of integration.

## The reverse chain rule

We can develop the ideas above to incorporate more complicated functions including some involving non-linear expressions raised to a power. The key to integrating these functions is recognising when an integral includes the product of an expression raised to a power and a multiple of its derivative.

Let's look at the following integral:

$\int\ 2x\left(x^2+3\right)^5\ dx$ 2x(x2+3)5 dx

Notice that, the derivative of the expression inside the brackets, $2x$2x, is also in the integral expression. As we did above, for a linear function raised to a power, we will solve this integral by considering a function with an index one higher than part of our integral expression.

The derivative of the function $f\left(x\right)=\left(x^2+3\right)^6$f(x)=(x2+3)6 with respect to $x$x is:

 $\frac{d}{dx}\left(x^2+3\right)^6$ddx​(x2+3)6 $=$= $6\times2x\left(x^2+3\right)^5$6×2x(x2+3)5 $=$= $12x\left(x^2+3\right)^5$12x(x2+3)5

Hence we could say that $12x\left(x^2+3\right)^5$12x(x2+3)5 is the derivative of $\left(x^2+3\right)^6$(x2+3)6 and $\left(x^2+3\right)^6$(x2+3)6 is an anti-derivative of $12x\left(x^2+3\right)^5$12x(x2+3)5.

We can use this result to explore the integral of $2x\left(x^2+3\right)^5$2x(x2+3)5:

 $\int\ 2x\left(x^2+3\right)^5\ dx$∫ 2x(x2+3)5 dx $=$= $\frac{1}{6}\int\ 12x\left(x^2+3\right)^5\ dx$16​∫ 12x(x2+3)5 dx

Notice that we already have the derivative of the term inside the brackets, $2x$2x. So we only have to multiply and divide the function to be integrated by the new index, in this case $6$6.

Using our previous result of $\frac{d}{dx}\left(x^2+3\right)^6=12x\left(x^2+3\right)^5$ddx(x2+3)6=12x(x2+3)5 and reversing the differentiation:

 $\frac{1}{6}\int\ 12x\left(x^2+3\right)^5\ dx$16​∫ 12x(x2+3)5 dx $=$= $\frac{1}{6}\left(x^2+3\right)^6+C$16​(x2+3)6+C

Notice that the derivative of the expression inside the brackets seemingly disappears in the integration process. Let's look at the reason for this by differentiating our integral to see if we end up with our original expression $2x(x^2+3)^5$2x(x2+3)5:

 $\frac{d}{dx}\frac{1}{6}\left(x^2+3\right)^6+C$ddx​16​(x2+3)6+C $=$= $\frac{1}{6}\times6\times2x\times\left(x^2+3\right)^5$16​×6×2x×(x2+3)5 $=$= $2x\left(x^2+3\right)^5$2x(x2+3)5

Hence, we can see that differentiating our integrated expression gives our original expression.

This rule is called the reverse chain rule and can be used in situations where the integrand is a product of a function raised to a power, and the function's derivative.

The reverse chain rule

$\int\ f'(x)\ f(x)^n\ dx=\frac{1}{n+1}f(x)^{n+1}+C$ f(x) f(x)n dx=1n+1f(x)n+1+C for $n\ne-1$n1

Note that there isn't an equivalent product or quotient rule in integration. Functions being integrated will often require algebraic manipulation into an integrable form prior to solving.

#### Worked examples

##### Example 4

Calculate $\int\ 4x^2\left(x^3+3\right)^7\ dx$ 4x2(x3+3)7 dx.

Think: The expression $\left(x^3+3\right)$(x3+3) has derivative $3x^2$3x2. We note that the $4x^2$4x2 term is a multiple of this derivative and therefore this is a candidate for the reverse chain rule. However, we will need to be careful to apply the correct factors to make the derivative and integral compatible.

Do: A strategy we can use is to rewrite the integral so that it has the exact derivative, and not just a multiple of the derivative. We will rewrite our integral as follows:

 $\int\ 4x^2\left(x^3+3\right)^7\ dx$∫ 4x2(x3+3)7 dx $=$= $\frac{4}{3}\int\ 3x^2\left(x^3+3\right)^7\ dx$43​∫ 3x2(x3+3)7 dx

We have multiplied by $\frac{4}{3}$43 to ensure we have not changed the integral overall. Now we can integrate this using the reverse chain rule as follows:

 $\frac{4}{3}\int\ 3x^2\left(x^3+3\right)^7\ dx$43​∫ 3x2(x3+3)7 dx $=$= $\frac{4}{3}\frac{\left(x^3+3\right)^8}{8}+C$43​(x3+3)88​+C $=$= $\frac{1}{6}\left(x^3+3\right)^8+C$16​(x3+3)8+C
##### Example 5

Calculate $\int\ \left(2x+1\right)\sqrt{x^2+x+1}\ dx$ (2x+1)x2+x+1 dx.

Think: The expression $\sqrt{x^2+x+1}$x2+x+1 can be rewritten as $\left(x^2+x+1\right)^{\frac{1}{2}}$(x2+x+1)12 and we note that $\left(2x+1\right)$(2x+1) is the derivative of the term under the radical sign. Hence, this is a candidate for the reverse chain rule.

Do: As our integral is in the form $\int\ f'\left(x\right)\ f(x)^n\ dx$ f(x) f(x)n dx we can use the reverse chain rule directly and write:

 $\int\ \left(2x+1\right)\sqrt{x^2+x+1}\ dx$∫ (2x+1)√x2+x+1 dx $=$= $\frac{2}{3}\left(x^2+x+1\right)^{\frac{3}{2}}+C$23​(x2+x+1)32​+C

#### Practice questions

##### Question 3

Find $\int8x^7\left(x^8-9\right)^6dx$8x7(x89)6dx, with $C$C as the constant of integration.

##### Question 4

Find $\int\left(2x+1\right)\left(x^2+x\right)^5dx$(2x+1)(x2+x)5dx, with $C$C as the constant of integration.

### Outcomes

#### 0606C14.9A

Integrate functions of the form (ax + b)^n for any rational n.