7.04 Antidifferentiation and logarithmic functions

The integral of the reciprocal function $y=\frac{1}{x}$y=1x​

Recall one of the first general rules of integration that we encountered:

$\int x^ndx=\frac{x^{n+1}}{n+1},n\ne-1$∫xndx=xn+1n+1​,n≠−1

In particular, notice that this rule does not apply for $n=-1$n=−1. Why is this?

Well, for $n=-1$n=−1, if we try to find $\int\frac{1}{x}dx$∫1x​dx using that rule we get the undefined solution $\frac{x^0}{0}$x00​

On the other hand, the graph of $y=\frac{1}{x}$y=1x​ most certainly has area between the curve and the $x$x-axis. So there should be a solution to definite integrals involving $\frac{1}{x}$1x​, as long as the interval does not cross $x=0$x=0 (where the function $y=\frac{1}{x}$y=1x​ is undefined).

So, what can we do about this apparent gap in the rule for integrating powers?

 

In our previous lesson we established that the function $y=\ln x$y=lnx has the derivative is given by $\frac{dy}{dx}=\frac{1}{x}$dydx​=1x​. This suggests that:

$\int\ \frac{1}{x}dx=\ln x+C$∫ 1x​dx=lnx+C

However, consider the following derivatives:

 

Original function	$f\left(x\right)=\ln\left(x\right)$f(x)=ln(x)		$f\left(x\right)=\ln\left(-x\right)$f(x)=ln(−x)		$f\left(x\right)=\ln\left(kx\right),k>0$f(x)=ln(kx),k>0
Derivative	$f'\left(x\right)$f′(x) $=$= $\frac{1}{x}$1x​		$f'\left(x\right)$f′(x) $=$= $\frac{-1}{-x}$−1−x​   $=$= $\frac{1}{x}$1x​		$f'\left(x\right)$f′(x) $=$= $\frac{k}{kx}$kkx​   $=$= $\frac{1}{x}$1x​
Derivative defined for	$x>0$x>0		$x<0$x<0		$x>0$x>0

So, when finding the anti-derivative of $y=\frac{1}{x}$y=1x​ how do we know which function to select as a primitive? The case where $f\left(x\right)=\ln\left(kx\right)$f(x)=ln(kx) (for $k>0$k>0), is in fact covered by $\int\ \frac{1}{x}dx=\ln x+C$∫ 1x​dx=lnx+C since $\ln(kx)$ln(kx) can be written as $\ln x+\ln k$lnx+lnk using our logarithm laws. However, without additional information restricting the domain of the function $y=\frac{1}{x}$y=1x​, the actual integral would be:

$\int\ \frac{1}{x}dx$∫ 1x​dx		$\ln x$lnx,	$x>0$x>0
		$\ln\left(-x\right)$ln(−x)	$x<0$x<0

 

Which can be stated as a single function by using the absolute value function:

$\int\ \frac{1}{x}\ dx=\ln\left(\left|x\right|\right)+C$∫ 1x​ dx=ln(|x|)+C

For this course and exercise we will only consider cases for $x>0$x>0, even when not explicitly stated, so we can omit the absolute value signs.

 

Practice questions

Question 1

Question 2

 

The integral of $y=\frac{1}{ax+b}$y=1ax+b​

For $y=\ln\left(ax+b\right)$y=ln(ax+b), the chain rule gives us a derivative of the form:

$\frac{dy}{dx}=\frac{a}{ax+b}$dydx​=aax+b​

Reversing this result, it follows that:

$\int\frac{1}{ax+b}dx=\frac{1}{a}\ln\left(\left|ax+b\right|\right)+C$∫1ax+b​dx=1a​ln(|ax+b|)+C

Again, for this course we will only consider cases where the argument of the resulting logarithmic function is positive.

 

Worked example

Example 1

Evaluate $\int\frac{12}{5x+1}dx$∫125x+1​dx

Think: The strategy here is to shift the factor $12$12 outside the integrand symbol and use $\int\frac{1}{ax+b}dx=\frac{1}{a}\ln\left(ax+b\right)+C$∫1ax+b​dx=1a​ln(ax+b)+C, where $a=5$a=5.

Do:

$\int\frac{12}{x+1}dx$∫12x+1​dx	$=$=	$12\int\frac{1}{5x+1}dx$12∫15x+1​dx
	$=$=	$\frac{12}{5}\ln\left(5x+1\right)+C$125​ln(5x+1)+C, where $C$C is a constant.

 

Practice questions

Question 3

Question 4

 

Further integrals resulting in logarithms

Beyond integrals of the forms examined above, we will approach further cases 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:

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

Let's look at an example using this approach.

 

Worked example

Example 2

Consider the function $y=\ln\left(x^3+7\right)$y=ln(x3+7).

(a) Find $\frac{dy}{dx}$dydx​.

Think: Use the fact that if $y=\ln f\left(x\right)$y=lnf(x), then $\frac{dy}{dx}=\frac{f'\left(x\right)}{f\left(x\right)}$dydx​=f′(x)f(x)​ which is an application of the chain rule.

Do:

$\frac{dy}{dx}$dydx​

$=$=

$\frac{3x^2}{x^3+7}$3x2x3+7​

 

(b) Hence, find the integral of $f(x)=\frac{12x^2}{x^3+7}$f(x)=12x2x3+7​, for $x^3+7>0$x3+7>0.

Think: Notice we have a multiple of the derivative found in part (a). Place the multiplier out the front and use $\int\frac{dy}{dx}dx=f\left(x\right)+C$∫dydx​dx=f(x)+C.

Do:

$\int\frac{12x^2}{x^3+7}\ dx$∫12x2x3+7​ dx	$=$=	$4\int\frac{3x^2}{x^3+7}\ dx$4∫3x2x3+7​ dx
	$=$=	$4\ln\left(x^3+7\right)+C$4ln(x3+7)+C, where $C$C is a constant.

 

Practice questions

Question 5

Question 6

 

Area examples

Consider the following examples carefully. The areas shown in each graph involves evaluating integrals of functions that lead to logarithmic functions.

 

Worked examples

Example 3

Find the area under the curve $y=\frac{16}{x-1}$y=16x−1​ between $x=3$x=3 and $x=9$x=9 .

Think: Sketching the function and area to be calculated assists with determining the best strategy for solving the problem. This can be done by hand or using technology.

The given function is a transformation of the basic hyperbola $y=\frac{1}{x}$y=1x​, by a translation $1$1 unit to the right and with a vertical dilation factor of $16$16. The vertical asymptote is given by $x=1$x=1 and the horizontal asymptote is given by $y=0$y=0. We could plot some points to obtain an accurate shape of the curve.

Do: The area is then given by:

$A$A	$=$=	$\int_3^9\frac{16}{x-1}dx$∫93​16x−1​dx
	$=$=	$16\left[\ln\left(x-1\right)\right]_3^9$16[ln(x−1)]93​
	$=$=	$16\left(\ln8-\ln2\right)$16(ln8−ln2)
	$=$=	$16\left(3\ln2-\ln2\right)$16(3ln2−ln2)
	$=$=	$32\ln2$32ln2

Hence, the area is $32\ln2$32ln2 units², which is approximately $22.18$22.18 units².

Example 4

Find the area under the curve $y=\frac{3}{6-4x}$y=36−4x​ between $x=0$x=0 and $x=1$x=1

Think: A sketch will help us visualise the area to be calculated. The hyperbola shown below has its vertical asymptote at $x=1.5$x=1.5 and a horizontal asymptote of $y=0$y=0. The graph has also been reflected and dilated.

Do: The area is then found as follows:

$A$A	$=$=	$\int_0^1\frac{3}{6-4x}dx$∫10​36−4x​dx
	$=$=	$-\frac{3}{4}\int_0^1\frac{-4}{6-4x}dx$−34​∫10​−46−4x​dx
	$=$=	$-\frac{3}{4}\left[\ln\left(6-4x\right)\right]_0^1$−34​[ln(6−4x)]10​
	$=$=	$-\frac{3}{4}\left(\ln2-\ln6\right)$−34​(ln2−ln6)
	$=$=	$\frac{3}{4}\left(\ln6-\ln2\right)$34​(ln6−ln2)
	$=$=	$\frac{3}{4}\ln3$34​ln3

Hence, the area is $\frac{3}{4}\ln3$34​ln3 units², which is approximately $0.824$0.824 units².

 

Practice questions

Question 7

Question 8