Example 8.1

A malfunctioning fridge maintains a temperature of 6°C which allows a population of bacteria to reproduce such that, on average, each bacteria divides every 20 min. Assuming no bacteria die in the time under consideration find a differential equation to describe the population growth.

Solution If the population at time t is given by p then the rate of change of the population is dp/dt. The increase in the population is such that it approximately doubles every 20 min, that is, it increases by p in 20 × 60 s. That gives a rate of increase as p/(20 × 60) per second. Hence, the differential equation describing the population is

$\frac{\text{d}p}{\text{d}t}=\frac{p}{1200}.$

Example 8.2

A capacitor, in an RC circuit, has been charged to a charge of Q₀. The voltage source has been removed and the circuit closed as in Figure 8.1. Find a differential equation that describes the rate of discharge of the capacitor if C = 0.001 μF and R = 10MΩ.

Solution The voltage across a capacitor is given by Q/C where C is the capacitance and Q is the charge on the capacitor. The voltage across the resistor is given by Ohm's Law as IR. From Kirchoff's voltage law, the sum of the voltage drops in the circuit must be 0; therefore, as the circuit is closed, we get voltage across the resistor + voltage across the capacitor = 0:

$\Rightarrow IR+\frac{Q}{C}=0.$

By definition, the current is the rate of change of charge with respect to time, that is I = d Q/dt giving the differential equation

$R\frac{\text{d}Q}{\text{d}t}+\frac{Q}{C}=0.$

We can rearrange this equation as

$R\frac{\text{d}Q}{\text{d}t}=-\frac{Q}{C}\iff \frac{\text{d}Q}{\text{d}t}=-\frac{Q}{RC}.$

We can see that this is an equation for exponential decay. The rate of change of the charge on the capacitor is proportional to the remaining charge at any point in time with a constant of proportionality given by 1/RC. In this case as R = 10MΩ and C = 0.001 μF, we get

$\frac{\text{d}Q}{\text{d}t}=-100Q.$

Example 8.3

Radioactivity is the emission of α- or, β-particles and γ-rays due to the disintegration of the nuclei of atoms. The rate of disintegration is proportional to the number of atoms at any point in time and the constant of proportionality is called the radioactivity decay constant. The radioactive decay constant for Radium B is approximately 4.3 × 10⁻⁴ s⁻¹. Give a differential equation that describes the decay of the number of particles N in a piece of Radium B.

Solution If the number of particles at time t is N, then the rate of change is dN/dt. The decay is proportional to the number of atoms, and we are given that the constant of proportionality is 4.3 × 10⁻⁴ s⁻¹ so that we have

$\frac{\text{d}N}{\text{d}t}=-4.3\times {10}^{-4}N$

as the equation which describes the decay.

Example 8.4

An object is heated so that its temperature is 400 K and the temperature of its surroundings is 300 K, and then it is left to cool. Newton's law of cooling states that the rate of heat loss is proportional to the excess temperature over the surroundings. Furthermore, if m is the mass of the object and c is its specific heat capacity then the rate of change of heat is proportional to the rate of fall of temperature of the body, and is given by

$\frac{\text{d}Q}{\text{d}t}=-mc\frac{\text{d}\varphi }{\text{d}t}$

where Q is the heat in the body and ϕ is its temperature. Find a differential equation for the temperature that describes the way the body cools.

Solution Newton's law of cooling gives

$\frac{\text{d}Q}{\text{d}t}=A(\varphi -{\varphi }_{\text{s}})$

where A is some constant of proportionality, Q is the heat in the body, ϕ is its temperature, and ϕ_s is the temperature of its surroundings. As we also know that

$\frac{\text{d}Q}{\text{d}t}=-mc\frac{\text{d}\varphi }{\text{d}t}$

this can be substituted in our first equation giving:

$-mc\frac{\text{d}\varphi }{\text{d}t}=A(\varphi -{\varphi }_{\text{s}})\iff \frac{\text{d}\varphi }{\text{d}t}=-\frac{A}{mc}(\varphi -{\varphi }_{\text{s}})$

A/(mc) can be replaced by a constant k, giving

$\frac{\text{d}\varphi }{\text{d}t}=-k(\varphi -{\varphi }_{\text{s}}).$

In this case, the temperature of the surroundings is known to be 300 K, so the equation describing the rate of change of temperature is

$\frac{\text{d}\varphi }{\text{d}t}=-k(\varphi -300).$

We have established that there are a number of important physical situations that can be described by the equation dy/dt = ky. The rate of change of y is proportional to its value. We would like to solve this equation, that is, find y explicitly as a function of t. In Chapter 7, we solved simple differential equations such as dy / dt = 3t by integrating both sides with respect to t, for example,

$\frac{\text{d}y}{\text{d}t}=3t\iff y=\int 3t?\text{d}t\iff y=\frac{3t^2}{2}+C.$

However, we cannot solve the equation dy/dt = ky in this way because the right-hand side is a function of y not of t. If we integrate both sides with respect to t we get

$\frac{\text{d}y}{\text{d}t}=ky\iff y=\int ky?\text{d}t.$

Although this is true we are no nearer solving for y as we need to know y as a function of t in order to find ∫ ky dt.

When we solved equations in Chapter 3 of the Background Mathematics notes available on the companion website for this book, we said that one method was to guess a solution and substitute that value for the unknown into the equation to see if it gave a true statement. This would be a very long method to use unless we are able to make an informed guess. We can use this method with this differential equation as we know from our experience with problems involving discrete growth that a solution should involve an exponential function of the form y = a^t. The main problem is to find the value of a, which will go with any particular equation. To do this we begin by looking for an exponential function that would solve the equation d y / dt = y, that is, we want to find the function whose derivative is equal to itself.

Example 8.5

(a) Show that any function of the form y = y₀e^t, where y₀ is a constant, is a solution to the equation

$\frac{\text{d}y}{\text{d}t}=y$

(b) Show that in the function y = yoe^t, y = y₀ when t = 0.

Solution

(a)  To show that y = y₀e^tare solutions, we first differentiate

$\frac{\text{d}}{\text{d}t}(y_0{\text{e}}^t)=y_0{\text{e}}^t$

(as y₀ is a constant and d (e^t)/dt = e^t).

Substitute for d y/dt and for y into the differential equation and we get y₀e^t= y₀e^t, which is a true statement for all t. Hence the solutions to dy/dt = y are y = y₀e^t.

(b) Substitute t = 0 in the function y = y₀e^t and we get y = y₀e^o. As any number raised to the power of 0 is 1, we have y = y0. Hence y₀ is the value of y at t = 0.

Using the function of a function rule we can find the derivative of e^kt, where k is some constant, and show that this function can be used to solve differential equations of the form dy/dt = ky.

Example 8.6

(a) Show that any function of the form y = y₀e^kt, where y₀ is a constant, is a solution to the equation

$\frac{\text{d}y}{\text{d}t}=ky$

(b) Show that in the function y = y₀e^kt, y = y₀ when t = 0.

Solution

(a) To show that y = y₀e^kt are solutions we first differentiate:

$\frac{\text{d}}{\text{d}t}(y_0{\text{e}}^{kt})=y_{0}k{\text{e}}^{kt}$

as y₀ is a constant and d(e^kt)/dt = ke^kt. Substitute for dy/dt and for y into the differential equation and we get

y₀ke^kt= ky₀e^ktwhich is a true statement for all values of t. Hence the solutions to dy /dt = ky are y = y₀e^kt.

(b) Substitute t = 0 in the function y = y₀e^tand we get y = y₀e⁰. As any number raised to the power of 0 is 1, we have y = y0. Hence, y₀ is the value of y at t = 0.

Example 8.7

Solve the differential equation given in Example 8.2, describing the discharge of a capacitor in a closed RC circuit with R = 10MΩ and C = 0.001 μF:

$\frac{\text{d}Q}{\text{d}t}=-100Q$

and find a particular solution given that at t = 0 the voltage drop against the capacitor was 1000 V.

Solution We have discovered that the solution to a differential equation of the form dy/dt = kt is given by y = y₀e^ktwhere y0 is the initial value of y.

Comparing dy/dt = ky with dQ/dt = −100Q, and replacing y by Q and k by −100, we get the solution

$Q=Q_0{\text{e}}^{-100t}$

To find the value of Q₀ we need to find the value of Q when t = 0. We are told that the initial value of the voltage across the capacitor was 1000 V and we know that the voltage drop across a capacitor is Q/C. Therefore, we have

$\begin{array}{lll}\frac{Q_0}{0.001\times {10}^{-6}}=1000\hfill & \iff \hfill & Q_0=1000\times 0.001\times {10}^{-6}\hfill \\ \hfill & \iff \hfill & Q_0={10}^{-6}\text{C}.\hfill \end{array}$

Therefore, the equation that describes the charge as the capacitor discharges is Q = 10^–6e^−100t C at time t s.

Example 8.8

Show that cosh(A + B) = cosh(A) cosh(B) + sinh(A) sinh(B).

Solution Substitute

$\begin{array}{l}cosh(A)=\frac{{\text{e}}^A+{\text{e}}^{-A}}{2}\\ sinh(A)=\frac{{\text{e}}^A-{\text{e}}^{-A}}{2}\\ cosh(B)=\frac{{\text{e}}^B+{\text{e}}^{-B}}{2}\\ sinh(B)=\frac{{\text{e}}^B+{\text{e}}^{-B}}{2}\end{array}$

into the right-hand side of the expression

cosh(A) cosh(B) + sinh(A) sinh(B)

$=\frac{({\text{e}}^A+{\text{e}}^{-A})}{2}\frac{({\text{e}}^B+{\text{e}}^{-B})}{2}\frac{({\text{e}}^A-{\text{e}}^{-A})}{2}\frac{({\text{e}}^B-{\text{e}}^{-B})}{2}.$

Multiplying out the brackets gives

$\frac{1}{4}\left({\text{e}}^{A+B}+{\text{e}}^{A-B}+{\text{e}}^{-A+B}+{\text{e}}^{-(A+B)}+(e^{A+B}-e^{A-B}-e^{-A+B}+e^{-(A+B)})\right).$

Simplifying then gives

$\frac{1}{4}(2{\text{e}}^{A+B}+2{\text{e}}^{-(A+B)})=\frac{1}{2}(e^{A+B}+e^{-(A+B)})$

which is the definition of cosh(A + B).

We have shown that the right-hand side of the expression is equal to the left-hand side, and therefore

cosh(A + B) = cosh(A) cosh(B) + sinh(A) sinh(B).

Example 8.9

Show that ${sinh}^{-1}(x)=ln(x+\sqrt{x^2+1})$ using the definitions

$y={sinh}^{-1}(x)????\iff ????sinh(y)=x$

and

$sinh(y)=\frac{{\text{e}}^y-{\text{e}}^{-y}}{2}$

Solution

$y={sinh}^{-1}(x)????\iff ????sinh(y)=x$

Using

$sinh(y)=\frac{{\text{e}}^y-{\text{e}}^{-y}}{2}$

to substitute on the left-hand side, we get

$\begin{array}{l}\frac{{\text{e}}^y-{\text{e}}^{-y}}{2}=x\\ ?????\iff ?????{\text{e}}^y-{\text{e}}^{-y}=2x????(\text{multiplying}?\text{by}?2)\\ ?????\iff ?????{\text{e}}^{2y}-1=2x{\text{e}}^y????(\text{multiplying}?\text{by}?{\text{e}}^y?\text{and}?\text{using}?\text{properties}?\text{of}\\ ???????????????????????????????????????????????????????\text{power}?\text{to}?\text{write}?{\text{e}}^y\cdot {\text{e}}^y={\text{e}}^{2y})\\ ?????\iff ?????{\text{e}}^{2y}-2x{\text{e}}^y-1=0?????(\text{subtracting}?2x{\text{e}}^y?\text{from}?\text{both}?\text{sides})\end{array}$

This is now a quadratic equation in e^y

${\left({\text{e}}^y\right)}^2-2x{\text{e}}^y-1=0.$

Using the formula for solving a quadratic equation, where a = 1, b = −2x, c = −1 gives

${\text{e}}^y=\frac{2x\pm \sqrt{4x^2+4}}{2}.$

Dividing the top and bottom lines by 2 gives

${\text{e}}^y=x\pm \sqrt{x^2+1}.$

Taking In of both sides and using In (e^y) = y (In is the inverse function of exp) we get

$y=ln(x\pm \sqrt{x^2+1}).$

We discount the negative sign inside the logarithm, as this would lead to a negative values, for which the logarithm is not defined. So finally

${sinh}^{-1}(x)=ln(x+\sqrt{x^2+1}).$

Example 8.10

Calculate the following, and where possible use the appropriate inverse function to check your result:

(a) sinh(1.444)

(b) tanh^−l (−0.5)

(c) cosh(−1)

(d) cosh^−l(3)

(e) cosh^−l(0)

Solution From the definition

$\begin{array}{c}sinh(1.444)=\frac{{\text{e}}^{1.444}-{\text{e}}^{-1.444}}{2}\\ \approx 2.0008152\\ =2.001?\text{to}?4?\text{s}.\text{f}.\end{array}$

Check: Use the inverse function of the sinh, that is, find sinh⁻¹ (2.0008152). From the logarithmic equivalence

(a) 

$\begin{array}{c}{sinh}^{-1}(2.0008152)=ln(2.0008152)+\sqrt{{(2.0008152)}^2+1}\\ =1.444\end{array}$

As this is the original number input into the sinh function we have found sinh⁻¹ (sinh(1.444)) = 1.444, which confirms the accuracy of our calculation.

(b) To calculate tanh −1 (−0.5) use the logarithmic equivalence giving

$\begin{array}{c}{tanh}^{-1}(0.5)=\frac{1}{2}ln\left(\frac{1+(-0.5)}{1-(-0.5)}\right)=\frac{1}{2}ln\left(\frac{1}{3}\right)\approx -0.5493061\\ =-0.5493?\text{to}?4\text{s}.\text{f}.\end{array}$

Check: Use the inverse function of tanh −1, that is, find

$tanh(-0.5493061)=\frac{{\text{e}}^{-0.5493061}-{\text{e}}^{0.5493061}}{{\text{e}}^{-0.5493061}+{\text{e}}^{0.5493061}}\approx -0.5$

This is the original number input to the tanh −1 function and this confirms the accuracy of our calculation.

(c) 

$\begin{array}{c}cosh(-1)=\frac{{\text{e}}^{-1}+{\text{e}}^1}{2}\approx 1.5430806\\ =1.543?\text{to}?4?\text{s}.\text{f}.\end{array}$

Check: Use the inverse function of cosh, that is, cosh −1 :

${cosh}^{-1}(1.5430806)=ln(1.5430806+\sqrt{{1.5430806}^2-1}\approx 1)$

This is not the number that we first started with, which was −1. However, we know that cosh⁻¹(x) is only a true inverse of cosh(x) if the domain of cosh(x) is limited to positive values and zero. We did not expect the inverse to ‘work’ in this case where we started with a negative value.

(d) To calculate cosh⁻¹ (3), use

$\begin{array}{c}{cosh}^{-1}(3)=ln(3+\sqrt{3^2-1})=ln(3+\sqrt{8})\\ \approx 1.7627472\\ =1.763?\text{to}?4?\text{s}.\text{f}.\end{array}$

Check: The inverse function to cosh ⁻¹ is cosh, so we find

$cosh(1.7627472)=\frac{{\text{e}}^{1.7627472}+{\text{e}}^{-1.7627472}}{2}\approx 3$

This confirms the accuracy of our calculation as we have shown cosh(cosh −1 (3)) = 3.

(e) Using the logarithmic definition of cosh ⁻¹ leads to an attempt to take the square root of a negative number. This confirms that cosh ⁻¹ (0) is not defined in .

Example 8.11

Show that$\frac{\text{d}}{\text{d}x}\left(sinh\left(x\right)\right)=cosh\left(x\right)$

Solution As

$sinh(x)=\frac{{\text{e}}^x-{\text{e}}^{-x}}{2}$

then

$\begin{array}{c}\frac{\text{d}}{\text{d}x}(sinh(x))=\frac{\text{d}}{\text{d}x}\left(\frac{{\text{e}}^x-{\text{e}}^{-x}}{2}\right)\\ =\frac{{\text{e}}^x-(-1){\text{e}}^{-x}}{2}=\frac{{\text{e}}^x+{\text{e}}^{-x}}{2}=cosh(x).\end{array}$

Therefore

$\frac{\text{d}}{\text{d}x}(sinh(x))=cosh(x).$

In a way similar to Example 8.6, we can find

$\frac{\text{d}}{\text{d}x}(cosh(x))=sinh(x).$

and

$\frac{\text{d}}{\text{d}x}(tanh(x))={sech}^2(x).$

The derivatives of the inverse hyperbolic functions can be found using the same method as given for the derivatives of the inverse trigonometric functions (in Chapter 5) and give

$\begin{array}{l}\frac{\text{d}}{\text{d}x}(sinh(x))=\frac{1}{\sqrt{1+x^2}}\\ \frac{\text{d}}{\text{d}x}({cosh}^{-1}(x))=\frac{1}{\sqrt{x^2-1}}\\ \frac{\text{d}}{\text{d}x}({tanh}^{-1}(x))=\frac{1}{1-x^2}.\end{array}$

Example 8.12

Find derivatives of the following:

$(\text{a})?y={\text{e}}^{-2t^2+3}????(\text{b})?x={\text{e}}^{-t}cos(3t)????(\text{c})y=\frac{sinh(x)}{x}??x\ne 0.$

Solution

(a) To differentiate y = e^−2t2+³using the function of a function rule think of this as y = e()(‘y = e to the bracket’).

Now differentiate y with respect to () and multiply by the derivative of () with respect to t. That is, use

$\frac{\text{d}y}{\text{d}t}=\frac{\text{d}y}{\text{d}()}\frac{\text{d}()}{\text{d}t}$

where () represents the expression in the bracket

$\frac{\text{d}y}{\text{d}t}={\text{e}}^{-2t^2+3}\frac{\text{d}}{\text{d}t}(-2t^2+3)={\text{e}}^{-2t^2+3}(-4t)=-4t{\text{e}}^{-2t^2+3}.$

(b) To find the derivative of x = e^−t cos (3t), write x = uv so that u = e^−t and v = cos(3t); then,

$\frac{\text{d}u}{\text{d}t}=-{\text{e}}^{-t}?????\frac{\text{d}\upsilon }{\text{d}t}=-3sin(3t)$

where we have used the chain rule to find both these derivatives. Now use the product rule

$\frac{\text{d}x}{\text{d}t}=u\frac{\text{d}\upsilon }{\text{d}t}+\upsilon \frac{\text{d}u}{\text{d}t}$

$\frac{\text{d}x}{\text{d}t}=-{\text{e}}^{-t}cos(3t)-{\text{e}}^{-t}3sin(3t)=-{\text{e}}^{-t}cos(3t)-3{\text{e}}^{-t}sin(3t).$

(c) To find the derivative of

$y=\frac{sinh\left(x\right)}{x}$

we use the formula for the quotient of two functions where y = u/v, u = sinh (x), v = x, and

$\frac{dy}{dx}=\frac{\nu \left(\text{d}u/\text{d}x\right)-u\left(\text{d}\nu /\text{d}x\right)}{{\nu }^2}$

Hence, we get

$\begin{array}{c}\frac{d}{dx}\left(\frac{sinh\left(x\right)}{x}\right)=\frac{xcosh\left(x\right)-sinh\left(x\right)\cdot 1}{x^2}\\ =\frac{x?cosh\left(x\right)-sinh\left(x\right)}{x^2}.\end{array}$

Example 8.13

Find the following integrals:

(a) 

${\displaystyle \int x{\text{e}}^{x^{\text{2}}\text{+2}}dx}$

(b) 

${\displaystyle \int \sinh (t)\text{?}{\cosh }^2(t)\text{?}dt}$

(c) 

${\displaystyle \int x{e}^x\text{?}\text{d}x}$

(d) 

${\displaystyle {\int }_1^2\ln (x)\text{d}x}$

(e) 

${\displaystyle \int \left(\frac{3x^2+2x}{x^3+x^2+2}\right)\text{d}x}$

Solution

(a) $\int x{\text{e}}^{x^2+2}\text{d}x$. Here, we have a function of a function e(x²+2) multiplied by a term that is something like the derivative of the term in the bracket.
Try a substitution, u = x²+ 2

$\Rightarrow \frac{\text{d}u}{\text{d}x}=2x\Rightarrow dx=\frac{du}{2x}$

then

$\int x{e}^{x^2+2}\text{d}x=\int x{e}^u\frac{\text{d}u}{2x}=\int \frac{1}{2}{e}^u\text{d}u=\frac{1}{2}{e}^u+C$

resubstituting u = x² + 2 gives

$\int x{e}^{x^2+2}dx=\frac{1}{2}{e}^{x^2+2}+C$

(b) To find ∫ sinh(t) cosh²(t) dt we remember that cosh²(t) = (cosh(t))2, so

$\int sinh\left(t\right){cosh}^2\left(t\right)dt=\int sinh\left(t\right){\left(cosh\left(t\right)\right)}^{2}dt$

sinh(t) is the derivative of the function in the bracket, cosh(t), so a substitution, u = cosh(t), should work:

$\begin{array}{c}u=cosh\left(t\right)\Rightarrow \frac{\text{d}u}{\text{d}t}=sinh\left(t\right)\\ \Rightarrow dt=\frac{du}{sinh\left(t\right)}\end{array}$

$\begin{array}{c}\int sinh\left(t\right){cosh}^2\left(t\right)\text{d}t=\int sinh\left(t\right)u^2\frac{du}{sinh\left(t\right)}=\int u^2\text{d}u\\ =u^3+C\end{array}$

resubstituting u = cosh(t) gives

$\int sinh\left(t\right)?{cosh}^2\left(t\right)dt=\frac{{cosh}^3\left(t\right)}{3}+C$

(c) ∫ xe^xdx. Use integration by parts

$\int u?\text{d}\nu =u\nu -\int \nu \text{d}u$

and choose u = x and dv = e^xdx giving

$\begin{array}{ccc}\frac{du}{dx}=1& \text{and}& \nu \int e^x\text{d}x=e^x\end{array}$

Then

$\begin{array}{c}\int x{e}^x\text{d}x=x{e}^x-\int e^x\text{d}x\\ =x{e}^x-e^x+C\end{array}$

(d) ${\int }_1^{2}ln\left(x\right)dx$. Write ln(x) = 1 ln (x) and use integration by parts with

$\begin{array}{ccc}u=ln\left(x\right)& \text{and}& d\nu =1?\text{d}x\end{array}$

$\begin{array}{cc}\Rightarrow \text{d}u=\frac{\text{d}x}{x}& \nu =\int 1?\text{d}x=x\end{array}$

$\begin{array}{c}{{\int }_1^{2}ln\left(x\right)dx=\left[x?ln\left(x\right)\right]}_1^2-{\int }_1^{2}x\frac{1}{2}dx\\ =2ln\left(2\right)-1ln\left(1\right)-{\int }_1^{2}1\text{d}x\\ =2ln\left(2\right)-{\left[x\right]}_1^2\\ =2ln\left(2\right)-\left(2-1\right)\approx 0.3863?\text{to}?4\text{s}.\text{f}.\end{array}$

(e) We rewrite

$\int \frac{3x^2+2x}{x^3+x^2+2}dx={\int \left(3x^2+2x\right)\left(x^3+x^2+2\right)}^{-1}dx.$

Notice that there are two brackets. To decide what to substitute we notice that

$\frac{d}{dx}\left(x^3+x^2+2\right)=3x^2+2x.$

so it should work to substitute $u=x^3+x^2+2$

$\Rightarrow \frac{d}{dx}=3x^2+2x\Rightarrow dx=\frac{dx}{3x^2+2x}.$

The integral becomes

$\begin{array}{c}\int \left(3x^2+2x\right)u^{-1}\frac{du}{3x^2+2x}\\ \int \frac{du}{u}=ln\left(u\right)+C.\end{array}$

Resubstituting for u gives

$\int \frac{3x^2+2x}{x^3+x^2+2}dx=ln\left(x^3+x^2+2\right)+C.$

Example 8.14

Find

(a) $\int \frac{2x-1}{\left(x-3\right)\left(x+1\right)}dx$

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

Solution

(a) $\int \frac{2x-1}{\left(x-3\right)\left(x+1\right)}dx.$

Rewrite the expression using partial fractions. We need to find A and B so that:

$\frac{2x-1}{\left(x-3\right)\left(x+1\right)}=\frac{A}{\left(x-3\right)}+\frac{B}{x+1}$

where this should be true for all values of x. Multiplying by (x – 3)(x + 1) gives 2x – 1 = A(x + 1) + B(x – 3). This is an identity, so we can substitute values for x: substitute x = −1 giving – 3 = B( −4) ⇔ B = 3/4 substitute x = 3 giving 5 = A(4) ⇔ A = 5/4

Hence,

$\frac{2x-1}{\left(x-3\right)\left(x+1\right)}=\frac{5}{4\left(x-3\right)}+\frac{3}{4\left(x+1\right)}.$

So

$\int \frac{2x-1}{\left(x-3\right)\left(x+1\right)}dx=\int \frac{5}{4\left(x-3\right)}+\frac{3}{4\left(x+1\right)}dx.$

As (x – 3) and (x + 1) are linear functions of x, we can find each part of this integral using substitutions of u = x – 3 and u = x + 1:

$\begin{array}{l}\int \frac{5}{4\left(x-3\right)}+\frac{3}{4\left(x+1\right)}dx=\frac{5}{4}ln\left(x-3\right)+\frac{3}{4}ln\left(x+1\right)+C\hfill \\ \int \frac{2x-1}{\left(x-3\right)\left(x+1\right)}dx=\frac{5}{4}ln\left(x-3\right)+\frac{3}{4}ln\left(x+1\right)+C.\hfill \end{array}$

Check:

$\frac{d}{dx}\left(\frac{5}{4}ln\left(x-3\right)+\frac{3}{4}ln\left(x+1\right)+C\right)=\frac{5}{4\left(x-3\right)}+\frac{3}{4\left(x+1\right)}$

writing this over a common denominator gives

$\begin{array}{r}\frac{d}{dx}\left(\frac{5}{4}ln\left(x-3\right)+\frac{3}{4}ln\left(x+1\right)+C\right)=\frac{5\left(x+1\right)+3\left(x-3\right)}{4\left(x-3\right)\left(x+1\right)}\hfill \\ =\frac{5x+5+3x-9}{4\left(x-3\right)\left(x+1\right)}\hfill \\ =\frac{8x-4}{4\left(x-3\right)\left(x+1\right)}\hfill \\ =\frac{2x-1}{\left(x-3\right)\left(x+1\right)}\hfill \end{array}$

(b) $\int \frac{x^2}{\left(x+2\right){\left(2x-1\right)}^2}dx$

Again, we can use partial fractions. Because of the repeated factor in the denominator we use both a linear and a squared term in that factor. We need to find A, B, and C so that

$\frac{x^2}{\left(x+2\right){\left(2x-1\right)}^2}=\frac{A}{\left(x+2\right)}+\frac{B}{\left(2x-1\right)}+\frac{C}{{\left(2x-1\right)}^2}$

where this should be true for all values of x. Multiply by (x + 2) (2x – 1)² to get

$x^2=A{(2x-1)}^2+B(2x-1)(x+2)+C(x+2)$

This is an identity, so we can substitute values for x

$\begin{array}{l}\text{substitute}?x=\frac{1}{2}?\text{giving}?0.25=C\left(2.5\right)\iff C=0.1\hfill \\ \text{substitute}?x=-2?\text{giving 4}?=\text{A}{\left(-5\right)}^2?\iff A=4/25=0.16\hfill \\ \text{substitute}?x=0?\text{giving}?\text{0}?=\text{A}+\text{B}\left(-1\right)\left(2\right)+C\left(2\right).\hfill \end{array}$

Using the fact that A = 0.16 and C = 0.1, we get

$0=0.16-2B+0.2\iff 0=0.36-2B\iff 2B=0.36\iff B=\mathrm{0.18.}$

Then we have

$\begin{array}{l}\int \frac{x^2}{\left(x+2\right){\left(2x-1\right)}^2}dx=\int \frac{0.16}{x+2}+\frac{0.18}{\left(2x-1\right)}+\frac{0.1}{{\left(2x-1\right)}^2}dx\hfill \\ =0.16ln\left(x+2\right)+\frac{0.18}{2}ln\left(2x-1\right)-\frac{0.1}{2}{\left(2x-1\right)}^{-1}+C\hfill \\ =0.16?ln\left(x+2\right)+0.09ln\left(2x-1\right)-\frac{0.05}{\left(2x-1\right)}+C.\hfill \end{array}$