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 $\int\frac{1}{x}dx$∫1xdx 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?
Recall that for the function $y=\ln x$y=lnx, we have established that the derivative is given by $\frac{dy}{dx}=\frac{1}{x}$dydx=1x. Using this, we can see that:
$\int\ \frac{1}{x}dx=\ln x+C$∫ 1xdx=lnx+C
In fact it is actually more correct to say that $\int\ \frac{1}{x}\ dx=\ln\left|x\right|+C$∫ 1x dx=ln|x|+C, since the logarithmic function is only defined for positive arguments while the reciprocal function is defined for all non-zero values of $x$x.
For the purposes of this discussion, however, we will omit the absolute value signs as is the regular practice of most school textbooks on this topic.
Thus we find that $\int_1^a\frac{1}{x}dx=\left[\ln x\right]_1^a=\ln a$∫a11xdx=[lnx]a1=lna as shown here:
$\int\frac{1}{x}dx=\ln\left|x\right|+C$∫1xdx=ln|x|+C, for some constant $C$C
The derivative of $f\left(x\right)$f(x) is $\frac{1}{x}$1x.
Which of the following could be the function? Select all the correct options.
$f\left(x\right)=k\ln x$f(x)=klnx
$f\left(x\right)=\ln\left(\left|kx\right|\right)$f(x)=ln(|kx|)
$f\left(x\right)=\ln kx$f(x)=lnkx for $k<0$k<0, $x>0$x>0
$f\left(x\right)=\ln x$f(x)=lnx
$f\left(x\right)=\ln kx$f(x)=lnkx for $k>0$k>0, $x<0$x<0
$f\left(x\right)=k\ln x$f(x)=klnx
$f\left(x\right)=\ln\left(\left|kx\right|\right)$f(x)=ln(|kx|)
$f\left(x\right)=\ln kx$f(x)=lnkx for $k<0$k<0, $x>0$x>0
$f\left(x\right)=\ln x$f(x)=lnx
$f\left(x\right)=\ln kx$f(x)=lnkx for $k>0$k>0, $x<0$x<0
Determine $\int\frac{3}{4x}dx$∫34xdx.
Use $C$C as the constant of integration.
Recall that we found that for $y=\ln\left(f\left(x\right)\right)$y=ln(f(x)), the chain rule gives us a derivative of the form $\frac{dy}{dx}=\frac{f'\left(x\right)}{f\left(x\right)}$dydx=f′(x)f(x).
Reversing this result, it follows that:
$\int\frac{f'\left(x\right)}{f\left(x\right)}dx=\ln\left|f\left(x\right)\right|+C$∫f′(x)f(x)dx=ln|f(x)|+C
That is, an expression of the form $\frac{f'\left(x\right)}{f\left(x\right)}$f′(x)f(x) (where the numerator is the derivative of the denominator) is the derivative of a logarithmic function.
$f\left(x\right)$f(x) can be any expression involving $x$x, so we can use this general result to find and/or evaluate many integrals. Let's look at a few examples.
Evaluate $\int\frac{12}{x+1}dx$∫12x+1dx
Think: The strategy here is to shift the factor $12$12 through the integrand symbol.
Do:
$\int\frac{12}{x+1}dx$∫12x+1dx | $=$= | $12\int\frac{1}{x+1}dx$12∫1x+1dx |
$=$= | $12\ln\left(x+1\right)+c$12ln(x+1)+c |
Note that since $\frac{\mathrm{d}}{\mathrm{d}x}\ln(x+1)=\frac{1}{x+1}$ddxln(x+1)=1x+1, then $\int\frac{dx}{x+1}=\ln(x+1)+c$∫dxx+1=ln(x+1)+c.
Evaluate $\int\frac{3x^2}{x^3+7}dx$∫3x2x3+7dx
Think: Here, if we imagine that $f(x)$f(x)$=x^3+7$=x3+7, then $f'(x)=3x^2$f′(x)=3x2, and so the integral is precisely in the form that is required. We can simply write down, without adjustment, the solution.
Do:
$\int\frac{3x^2}{x^3+7}dx$∫3x2x3+7dx | $=$= | $\ln(x^3+7)+c$ln(x3+7)+c |
Find the exact value of $\int_4^6\frac{1}{x-2}dx$∫641x−2dx.
Express your answer in simplest form.
Determine $\int\frac{1}{2x+3}dx$∫12x+3dx.
You may use $C$C as the constant of integration.
Consider $\int\frac{1}{2x-3}dx$∫12x−3dx.
In this case, the derivative of the denominator is $2$2. So if the expression were $\int\frac{2}{2x-3}dx$∫22x−3dx, we could immediately recognise it as the derivative of $\ln\left(2x-3\right)+C$ln(2x−3)+C.
With a bit of manipulation, we can rewrite the integrand in the form we want:
$\int\frac{1}{2x-3}dx$∫12x−3dx | $=$= | $\int\frac{1}{2}\times\frac{2}{2x-3}dx$∫12×22x−3dx |
$=$= | $\frac{1}{2}\int\frac{2}{2x-3}dx$12∫22x−3dx |
By rewriting the expression in this slightly modified form, we get the integral $\frac{f'\left(x\right)}{f\left(x\right)}$f′(x)f(x), which we then know integrates to give $\frac{1}{2}\ln\left(2x-3\right)+C$12ln(2x−3)+C.
So if the integrand is not immediately in the form we want, we can see if multiplying and dividing by constants will help rewrite it in the form of the derivative of a log function.
Things to consider
In deciding whether or not logarithms are involved in the evaluation of an integral, we should first look to see if the integrand (the function being integrated) is given as an algebraic fraction. If so, we should check the following:
Evaluate $\int_1^3\frac{12x^2+4x+1}{2x}dx$∫3112x2+4x+12xdx
Think: The strategy is to split the integrand up into three parts and integrate term by term.
Do:
$\int_1^3\frac{12x^2+4x+1}{2x}dx$∫3112x2+4x+12xdx | $=$= | $\int_1^36x+2+\frac{1}{2x}dx$∫316x+2+12xdx |
$=$= | $\left[\frac{6x^2}{2}+2x+\frac{1}{2}\ln x\right]_1^3$[6x22+2x+12lnx]31 | |
$=$= | $\left[\left(27+6+\frac{1}{2}\ln3\right)-\left(3+2+0\right)\right]$[(27+6+12ln3)−(3+2+0)] | |
$=$= | $28+\frac{1}{2}\ln3$28+12ln3 |
Evaluate $\int\frac{6x^2-1}{4x^3-2x+1}dx$∫6x2−14x3−2x+1dx
Think: The derivative of the denominator function is a factor of $2$2 more than the numerator function. The strategy is then to increase the numerator by a factor of $2$2, but then reduce the integral by a factor of a half to restore the original question.
Do:
$\int\frac{6x^2-1}{4x^3-2x+1}dx$∫6x2−14x3−2x+1dx | $=$= | $\frac{1}{2}\int\frac{12x^2-2}{4x^3-2x+1}dx$12∫12x2−24x3−2x+1dx |
$=$= | $\frac{1}{2}\ln\left(4x^3-2x+1\right)+C$12ln(4x3−2x+1)+C |
$\int\frac{f'\left(x\right)}{f\left(x\right)}dx=\ln\left|f\left(x\right)\right|+C$∫f′(x)f(x)dx=ln|f(x)|+C, for some constant $C$C
Determine $\int\frac{4x-4}{x^2-2x}dx$∫4x−4x2−2xdx.
You may use $C$C as the constant of integration.
The gradient of a curve at any point is given by $f'\left(x\right)=\frac{5}{5x-4}$f′(x)=55x−4, and the curve passes through the point $\left(3,\ln11\right)$(3,ln11).
Define $f\left(x\right)$f(x). You may use $C$C to represent the constant of integration if needed.
Consider the following examples carefully. The areas shown in each graph involves evaluating integrals of functions that lead directly to log functions.
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.
Do: The sketch is quite straight forward - a hyperbola translated $1$1 unit to the right with a dilation factor of $16$16. We could find some points on the curve by thinking about certain values of $x$x - for example, at $x=5$x=5 we have $y=\frac{16}{5-1}=4$y=165−1=4, and at $x=9$x=9, $y=\frac{16}{9-1}=2$y=169−1=2. The vertical asymptote is given by $x=1$x=1.
The area is then given by:
$A$A | $=$= | $\int_3^9\frac{16}{x-1}dx$∫9316x−1dx |
$=$= | $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 | |
$\approx$≈ | $22.18$22.18 |
Hence the area is approximately $22.18$22.18 units^{2}.
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.
Do: The inverted hyperbola shown below has its vertical asymptote at $x=1.5$x=1.5 and a dilation factor of $3$3.
The area is then found as follows:
$A$A | $=$= | $\int_0^1\frac{3}{6-4x}dx$∫1036−4xdx |
$=$= | $-\frac{3}{4}\int_0^1\frac{-4}{6-4x}dx$−34∫10−46−4xdx | |
$=$= | $-\frac{3}{4}\left[\ln\left(6-4x\right)\right]_0^1$−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) | |
$\approx$≈ | $0.824$0.824 |
Consider the function $y=\frac{2}{3x+5}$y=23x+5.
Sketch a graph of the function.
Find the area enclosed by the curve, the coordinate axes and the line $x=4$x=4.
Find the area enclosed by the function $f\left(x\right)=\frac{x}{x^2+2}$f(x)=xx2+2, the $x$x-axis and the lines $x=2$x=2 and $x=4$x=4.
Integrate sums of terms in powers of x including 1/x and 1/(ax +b).
Integrate functions of the form e^(ax + b).