7.05 Further anti-differentiation

Integration of $f\left(ax+b\right)$f(ax+b)

In previous lessons we have seen the following rules for integration:

Function $f\left(x\right)$f(x)	Integral $\int f\left(x\right)dx$∫f(x)dx
$e^{ax+b}$eax+b	$\frac{1}{a}e^{ax+b}+C$1a​eax+b+C
$\cos\left(ax+b\right)$cos(ax+b)	$\frac{1}{a}\sin\left(ax+b\right)+C$1a​sin(ax+b)+C
$\sin\left(ax+b\right)$sin(ax+b)	$-\frac{1}{a}\cos\left(ax+b\right)+C$−1a​cos(ax+b)+C

In each case we had a function composed with a linear function, $ax+b$ax+b. When differentiating such a function we would use the chain rule and multiply the derivative of the function as a whole by the derivative of the inside function (in this case $a$a). Reversing the process involved dividing the integral by the constant $a$a.

A form of this integral we have not yet looked at is where $f\left(x\right)=\left(ax+b\right)^n$f(x)=(ax+b)n. That is, where we have a linear function raised to a power.

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

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

For example:

$\frac{d}{dx}\left(2x+5\right)^3$ddx​(2x+5)3	$=$=	$3\times\left(2x+5\right)^{3-1}\times2$3×(2x+5)3−1×2
	$=$=	$6(2x+5)^2$6(2x+5)2

 

We have established that integration is the reverse of differentiation. Thus, reversing the process above we obtain the following steps to integrate a linear function raised to a power:

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

For example:

$\int\left(2x+5\right)^2dx$∫(2x+5)2dx	$=$=	$\frac{\left(2x+5\right)^{2+1}}{2\times3}$(2x+5)2+12×3​
	$=$=	$\frac{1}{6}\left(2x+5\right)^3$16​(2x+5)3

 

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

 

Worked examples

Example 1

Find $\int(2x+1)^3dx$∫(2x+1)3dx.

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.

$\int(2x+1)^3dx$∫(2x+1)3dx	$=$=	$\frac{(2x+1)^4}{4\times2}+C$(2x+1)44×2​+C
	$=$=	$\frac{(2x+1)^4}{8}+C$(2x+1)48​+C, where $C$C is a constant

Example 2

Let  $f\left(x\right)=\frac{2}{(5-x)^4}$f(x)=2(5−x)4​. Find an antiderivative of $f\left(x\right)$f(x).

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 rewrite this expression using negative indices we can get it into a form that we can integrate.

Do:

$\int\frac{2}{\left(5-x\right)^4}dx$∫2(5−x)4​dx	$=$=	$\int2\left(5-x\right)^{-4}dx$∫2(5−x)−4dx	Rewrite the expression using negative powers
	$=$=	$\frac{2\left(5-x\right)^{-4+1}}{-1\times-3}+C$2(5−x)−4+1−1×−3​+C	Add one to the power, then divide by the new power and derivative of the linear function
	$=$=	$\frac{2\left(5-x\right)^{-3}}{3}+C$2(5−x)−33​+C	Simplify
	$=$=	$\frac{2}{3\left(5-x\right)^3}+C$23(5−x)3​+C, where $C$C is a constant	Rewrite with positive powers

 

As the questions asks for an antiderivative, a particular solution must be given.  For example, an antiderivative is $\frac{2}{3\left(5-x\right)^3}$23(5−x)3​.

Practice questions

Question 1

Question 2

Question 3

 

Further anti-differentiation strategies

In each case above we were looking at functions composed with a linear function. What about integrating functions that are composed with more complex functions, such as $\left(x^4+5x\right)^2$(x4+5x)2, $\left(3x^2-2x+1\right)^4$(3x2−2x+1)4, $xe^{3x^2}$xe3x2, or $x^2\sin(4x^3+5)$x2sin(4x3+5)? In the first case we could expand the brackets before integrating, but this strategy is limited to low powers. For instance, we can see that to expand the second case would be very tedious, and furthermore this strategy is of no use in the other two cases as they are not just power functions.

While there are more advanced techniques to integrate such functions, we will approach these by differentiating a related function and looking to manipulate the integral into a form that uses the property that differentiation and integration are reverse operations. That is, if we have a function $y=f\left(x\right)$y=f(x), then:

$\int\frac{dy}{dx}dx=f(x)+C$∫dydx​dx=f(x)+C, where $C$C is a constant

Let's look at some examples using this approach.

 

Worked examples

Example 3

Consider the function $y=\left(x^2+10\right)^4$y=(x2+10)4.

(a) Find $\frac{dy}{dx}$dydx​.

Think: We have a function to a power. Hence, we will use the rule $\frac{d}{dx}\left[f\left(x\right)\right]^n=nf'\left(x\right)\left[f\left(x\right)\right]^{n-1}$ddx​[f(x)]n=nf′(x)[f(x)]n−1.

Do:

$\frac{d}{dx}\left(x^2+10\right)^4$ddx​(x2+10)4	$=$=	$4\times2x\left(x^2+10\right)^{4-1}$4×2x(x2+10)4−1
	$=$=	$8x\left(x^2+10\right)^3$8x(x2+10)3

 

(b) Hence, determine $\int32x\left(x^2+10\right)^3dx$∫32x(x2+10)3dx.

Think: Since integration is the reverse of differentiation, part (a) tells us that $\int8x\left(x^2+10\right)^3dx=\left(x^2+10\right)^4+C$∫8x(x2+10)3dx=(x2+10)4+C, for some constant $C$C.

Notice the expression that we are trying to integrate is very similar to this, but has a different constant at the front. So in order to integrate the expression, let's first rewrite the constant $32$32 as a multiple of $8$8.

Observe that $32=4×8$32=4×8. We can use the fact that $\int kf\left(x\right)dx=k\int f\left(x\right)dx$∫kf(x)dx=k∫f(x)dx to write this extra factor of $4$4 in front of the integral.

Do:

$\int32x\left(x^2+10\right)^3dx$∫32x(x2+10)3dx	$=$=	$\int4\times8x\left(x^2+10\right)^3dx$∫4×8x(x2+10)3dx	Rewrite $32$32 as a multiple of $8$8
	$=$=	$4\int8x\left(x^2+10\right)^3dx$4∫8x(x2+10)3dx	Bring the factor of $4$4 to the front and use the information from part (a) to integrate
	$=$=	$4\left(x^2+10\right)^4+C$4(x2+10)4+C, where $C$C is a constant

 

Example 4

Given $f(x)=xe^x$f(x)=xex.

(a) Find $f'(x)$f′(x).

Think: Here the function is of the form of the product of two functions, so we will use the product rule.

Do:

Let	$u=x$u=x	then	$u'=1$u′=1
and	$v=e^x$v=ex	then	$v'=e^x$v′=ex

$f'\left(x\right)$f′(x)	$=$=	$uv'+vu'$uv′+vu′	Write out the product rule
	$=$=	$x\times e^x+e^x\times1$x×ex+ex×1	Make the appropriate substitutions
	$=$=	$xe^x+e^x$xex+ex	Simplify

 

(b) Hence, determine the integral $\int xe^xdx$∫xexdx.

Think: Since integration is the reverse of differentiation, part (a) tells us that $\int xe^x+e^xdx=xe^x+C$∫xex+exdx=xex+C, for some constant $C$C.

We have been asked to determine the integral of the first term of the function inside the integral so we will use the property $\int f\left(x\right)+g\left(x\right)dx=\int f\left(x\right)dx+\int g\left(x\right)dx$∫f(x)+g(x)dx=∫f(x)dx+∫g(x)dx to split the integral and then rearrange to find the part we require.

Do:

$\int xe^x+e^xdx$∫xex+exdx	$=$=	$xe^x+C$xex+C	From part (a)
$\int xe^xdx+\int e^xdx$∫xexdx+∫exdx	$=$=	$xe^x+C$xex+C	Using the property $\int f\left(x\right)+g\left(x\right)dx=\int f\left(x\right)dx+\int g\left(x\right)dx$∫f(x)+g(x)dx=∫f(x)dx+∫g(x)dx
$\int xe^xdx+e^x$∫xexdx+ex	$=$=	$xe^x+C$xex+C	Evaluate $\int e^xdx$∫exdx
$\therefore\int xe^xdx$∴∫xexdx	$=$=	$xe^x-e^x+C$xex−ex+C, where $C$C is a constant	Rearrange

 

Practice questions

Question 4

Question 5

Question 6