9.1 Workshop Solutions 1

True or False? (i) The answer depends on whether we count roots with multiplicity or not. For example, if $p(x)=(x-1)^{2}(x^{2}+1)$ then $p$ has degree 4 and has one real root with multiplicity two. If we only count the number of distinct roots then $n=4$, $s=1$ and $n-s=3$ is odd. If we consider $p(x)$ to have two real roots (which happen to be equal) then $n=4$, $s=2$ and $n-s=2$ is even.

In the general case, $p(x)=(x-a_{1})\ldots(x-a_{s})q_{1}(x)\ldots q_{j}(x)$ as in slide 1.6 in the notes, and $n=s+2j$ so $n-s$ is even.

(ii) False. For example, $(x-1)^{3}$ has degree 3 but only one root (with multiplicity 3).

(iii) True. We complete the square, but first divide through by $(-2)$ to obtain $x^{2}-3x+\frac{7}{2}$, which equals $(x-\frac{3}{2})^{2}+\frac{5}{4}$.

(iv) False. We have $1-x^{2}=\cos^{2}t$ so $\sqrt{1-x^{2}}=\sqrt{\cos^{2}t}=|\cos t|$. For $\frac{\pi}{2}\leq t\leq\pi$, $\cos t$ is negative so $|\cos t|=-\cos t$ in that range. Therefore $\sqrt{1-x^{2}}=\left\{\begin{array}[]{cc}\cos t&0\leq t\leq\frac{\pi}{2},\\ -\cos t&\frac{\pi}{2}\leq t\leq\pi.\end{array}\right.$

W1.1 i) $\frac{x-2}{x+4}=\frac{x+4-6}{x+4}=1-\frac{6}{x+4}$.

ii) $\frac{x^{2}+x}{x-3}=\frac{x(x-3)+4x}{x-3}=\frac{x(x-3)+4(x-3)+12}{x-3}=x+4+\frac{12}{x-3}$.

iii) $\frac{x^{2}-1}{x^{2}+x+3}=\frac{x^{2}+x+3-(x+4)}{x^{2}+x+3}=1-\frac{x+4}{x^{2}+x+3}$.

iv) $\frac{x^{3}-2x-1}{x^{2}+2x+2}=\frac{x(x^{2}+2x+2)-2x^{2}-4x-1}{x^{2}+2x+2}=x+\frac{-2(x^{2}+2x+2)+3}{x^{2}+2x+2}=x-2+\frac{3}{x^{2}+2x+2}$.

W1.2 i) Here the degree of the numerator is less than the degree of the denominator, so we don’t require a polynomial term. We have $x^{2}-x=x(x-1)$, so the required form of partial fractions is $\frac{A}{x}+\frac{B}{x-1}$. Then

$$\frac{A}{x}+\frac{B}{x-1}=\frac{5x-3}{x(x-1)}\;\Leftrightarrow\;A(x-1)+Bx=5x-3.$$

In this case we have the easy trick of setting $x=1$ to determine $B$ and $x=0$ to find $A$. When $x=1$ we have $B=2$, and when $x=0$ we have $-A=-3$, so $A=3$.

ii) Once again we don’t require a polynomial term. We factorize $x^{2}-x-2=(x-2)(x+1)$ so the required form of partial fractions is $\frac{A}{x-2}+\frac{B}{x+1}$. Then

$$\frac{6x-9}{x^{2}-x-2}=\frac{A}{x-2}+\frac{B}{x+1}\;\Leftrightarrow\;A(x+1)+B(x-2)=6x-9\;\Leftrightarrow A=1,B=5.$$

iii) Again we don’t require a polynomial term. We factorize $g(x)=x^{3}+4x^{2}+5x+2$ by looking for roots. We have $g(-1)=-1+4-5+2=0$ so $g(x)$ is divisible by $x+1$. Now $g(x)=(x+1)(x^{2}+3x^{2}+2x)$. Finally, $x^{2}+3x+2x=(x+1)(x+2)$ so $g(x)=(x+1)^{2}(x+2)$. Hence the required form of partial fractions is $\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{x+2}$. Then

$$\frac{x^{2}+5x+5}{x^{3}+4x^{2}+5x+2}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{x+2}\;\Leftrightarrow\;x^{2}+5x+5=A(x+1)(x+2)+B(x+2)+C(x+1)^{2}.$$

We can carry out the trick of setting $x=-1$ and $x=-2$ to determine $B$ and $C$, but this won’t be enough to tell us the value of $A$. Setting $x=-1$ we obtain $1=B$, and setting $x=-2$ we have $-1=C$. Now we can subtract $x+2$ from the left and right to get $x^{2}+4x+3=A(x+1)(x+2)-(x+1)^{2}$, and then we add $(x+1)^{2}$ to the left and right to get $2x^{2}+6x+4=A(x+1)(x+2)=A(x^{2}+3x+2)$, whence $A=2$. Thus $\frac{x^{2}+5x+5}{x^{3}+4x^{2}+5x+2}=\frac{2}{x+1}+\frac{1}{(x+1)^{2}}-\frac{1}{x+2}$.

iv) In this case $\deg f=1<\deg 3=3$, so there is no need to carry out polynomial long division. We first have to factorize the denominator. We observe that $g(2)=8-8+8-8=0$, so $g(x)$ is divisible by $(x-2)$. We have $g(x)=(x-2)(x^{2}+4)$.

To carry out the partial fractions step, we want to express $f(x)/g(x)$ as $\frac{A}{x-2}+\frac{Bx+C}{x^{2}+4}$. Multiplying through by $g(x)$, we have $A(x^{2}+4)+(Bx+C)(x-2)=2x-20$. We can evaluate at $x=2$ to obtain $8A=-16$ and hence $A=-2$. However, it isn’t a good idea to try to obtain the coefficients $B,C$ by evaluating at $\pm 2{\rm i}$. There are two ways of finding $B,C$. The standard approach is to collect the terms on the left-hand side to obtain: $(A+B)x^{2}+(C-2B)x+(4A-2C)=2x-20$. Then $A+B=0$, so $B=-A=2$, and $C-2B=2$, so $C=2+2B=6$.

An alternative “trick” method is to substract $A(x^{2}+4)$ from both sides of the equation to obtain: $(Bx+C)(x-2)=2x-20+2(x^{2}+4)=2x^{2}+2x-12$. Then $2x^{2}+2x-12=2(x^{2}+x-6)=2(x+3)(x-2)$ and hence, dividing both sides by $(x-2)$, we obtain $Bx+C=2x+6$.

The answer (by either method) is therefore $\frac{2x+6}{x^{2}+4}-\frac{2}{x-2}$.

v) Here $\deg f=\deg g$, so we need to carry out polynomial division before the partial fractions step. We have $x^{3}+3x^{2}-x-14=x^{3}+3x^{2}+2x-6-(3x+8)$, so $\frac{x^{3}+3x^{2}-x-14}{x^{3}+3x^{2}+2x-6}=1-\frac{3x+8}{x^{3}+3x^{2}+2x-6}$. Now we factorize $g(x)=x^{3}+3x^{2}+2x-6$. We have $g(1)=0$ so $g(x)$ is divisible by $(x-1)$. Then $g(x)=(x-1)(x^{2}+4x+6)$. We now observe that $x^{2}+4x+6=(x+2)^{2}+2$ is irreducible. Thus the required form of partial fractions is $\frac{A}{x-1}+\frac{Bx+C}{x^{2}+4x+6}$. Then

$$\frac{3x+8}{x^{3}+3x^{2}+2x-6}=\frac{A}{x-1}+\frac{Bx+C}{x^{2}+4x+6}\;\Leftrightarrow\;3x+8=A(x^{2}+4x+6)+(Bx+C)(x-1).$$

Now we can carry out the trick of setting $x=1$ to obtain $11=11A$ and hence $A=1$. Using the standard method (equating coefficients) to find $B,C$, we have $3x+8=(A+B)x^{2}+(4A+C-B)x+(6A-C)$, and hence $A+B=0$, $4A+C-B=3$ and $6A-C=8$. Since $A=1$, we obtain $B=-1$ and hence $C=3-4A+B=-2$. With the trick method, we subtract $x^{2}+4x+6$ from both sides to obtain $-x^{2}-x-2=(Bx+C)(x-1)$ and hence, dividing by $(x-1)$, we have $Bx+C=-x-2$.

The answer to the question is therefore $1-\frac{1}{x-1}+\frac{x+2}{x^{2}+4x+6}$.

W1.3 i) We have $\frac{x+3}{(x-1)^{2}}=\frac{x-1+4}{(x-1)^{2}}=\frac{1}{x-1}+\frac{4}{(x-1)^{2}}$. Hence the integral is $\log|x-1|-\frac{4}{x-1}+c$.

ii) We make the substitution $x=3\tan t$, so that $\frac{dx}{dt}=3\sec^{2}t$, and $x^{2}+9=9(1+\tan^{2}t)=9\sec^{2}t$. So

$$\int\frac{dx}{x^{2}+9}=\int\frac{\frac{dx}{dt}\,dt}{9\sec^{2}t}=\frac{1}{3}\int\,dt=\frac{t}{3}+c=\frac{1}{3}\tan^{-1}\frac{x}{3}+c.$$

(Alternatively, we can just remember the fact that $\int\frac{dx}{x^{2}+a^{2}}=\frac{1}{a}\tan^{-1}\frac{x}{a}+c$.)

iii) We separate the numerator into the $x$ term and the constant term, so we have to integrate $\frac{4x}{x^{2}+1}$ and $\frac{9}{x^{2}+1}$ separately. For the first term we make the substitution $u=x^{2}+1$, so that $\frac{du}{dx}=2x$ and so we have $\int\frac{4x\,dx}{x^{2}+1}=\int\frac{2\,du}{u}=2\log|u|+c=2\log(x^{2}+1)+c$. The second term is $9\tan^{-1}x$. (This is more or less standard, or one can make the substitution $x=\tan t$.)

iv) Once again we separate the numerator, so we have to separately integrate $\frac{x}{x^{2}+3}$ and $\frac{2}{x^{2}+3}$. For the first term, we substitute $u=x^{2}+3$, to obtain $\int\frac{x\,dx}{x^{2}+3}=\frac{1}{2}\log(x^{2}+3)+c$. For the second term, we substitute $x=\sqrt{3}\tan t$, or we remember the integral of $\frac{1}{x^{2}+a^{2}}$ with $a=\sqrt{3}$. Then we obtain $\int\frac{2}{x^{2}+3}=\frac{2}{\sqrt{3}}\tan^{-1}\frac{x}{\sqrt{3}}+c^{\prime}$. Then

$$\int\frac{x+2}{x^{2}+3}\,dx=\frac{1}{2}\log(x^{2}+3)+\frac{2}{\sqrt{3}}\tan^{-1}\frac{x}{\sqrt{3}}+c.$$

W1.4 i) By our answer to W1.2(i), we have $\int\frac{5x-3}{x^{2}-x}\,dx=\int\frac{3\,dx}{x}+\int\frac{2\,dx}{x-1}=3\log|x|+2\log|x-1|+c=\log(|x^{3}(x-1)^{2}|)+c$.

ii) Using our answer above, we have $\int\frac{6x-9}{x^{2}-x-2}dx=\int\frac{dx}{x-2}+\int\frac{5\,dx}{x+1}=\log|x-2|+5\log|x+1|+c=\log|(x-2)(x+1)^{5}|+c$

iii) We have $\int\frac{x^{2}+5x+5}{x^{3}+4x^{2}+5x+2}\,dx=\int\frac{2\,dx}{x+1}+\int\frac{dx}{(x+1)^{2}}-\int\frac{dx}{x+2}=2\log|x+1|-\frac{1}{x+1}-\log|x+2|+c=\log\left|\frac{(x+1)^{2}}{x+2}\right|-\frac{1}{x+1}+c$.

iv) By our answer to 1.2(iv), we have $\int\frac{2x-20}{x^{3}-2x^{2}+4x-8}\,dx=\int\frac{2x+6}{x^{2}+4}\,dx-\frac{2\,dx}{x-2}$. To integrate $\frac{2x}{x^{2}+4}$ we substitute $u=x^{2}+4$, hence $\frac{du}{dx}=2x$ and so $\int\frac{2x\,dx}{x^{2}+4}=\int\frac{du}{u}=\log|u|+c=\log(x^{2}+4)+c$. To integrate $\frac{6}{x^{2}+4}$, we substitute $x=2\tan t$, hence $\frac{dx}{dt}=2\sec^{2}t$ and $x^{2}+4=4(\tan^{2}t+1)=4\sec^{2}t$. Thus $\int\frac{6\,dx}{x^{2}+4}=\int\frac{12\sec^{2}t\,dt}{4\sec^{2}t}=3\int\,dt=3t+c^{\prime}=3\tan^{-1}\left(\frac{x}{2}\right)+c^{\prime}$. Finally, we can immediately integrate $\frac{2}{x-2}$ to obtain $2\log|x-2|+c^{\prime\prime}$. Summing all of the terms together, we have:

$$\int\frac{2x-20}{x^{3}-2x^{2}+4x-8}\,dx=\log(x^{2}+4)+3\tan^{-1}\left(\frac{x}{2}\right)-2\log|x-2|+C.$$

v) We have $\int\frac{x^{3}+3x^{2}-x-14}{x^{3}+3x^{2}+2x-6}dx=\int 1\,dx-\int\frac{dx}{x-1}+\int\frac{x+2}{x^{2}+4x+6}\,dx=x-\log|x-1|+\int\frac{x+2}{x^{2}+4x+6}dx$. To obtain the final integral, we substitute $u=x^{2}+4x+6$. Then $\frac{du}{dx}=2x+4$ and therefore $\int\frac{x+2}{x^{2}+4x+6}dx=\frac{1}{2}\int\frac{du}{u}=\log|u|+c$. Then

$$\int\frac{x^{3}+3x^{2}-x-14}{x^{3}+3x^{2}+2x-6}dx=x-\log|x-1|+\frac{1}{2}\log(x^{2}+4x+6)+c=x+\frac{1}{2}\log\left(\frac{x^{2}+4x+6}{(x-1)^{2}}\right)+c.$$

W1.5. i) First we factorize the denominator $x^{3}-x^{2}-8x+12$. By considering factors of $12$, we spot $2$ as a root, and then proceed to factorize

$${x^{3}-x^{2}-8x+12=(x-2)(x^{2}+x-6)=(x-2)(x-2)(x+3)=(x-2)^{2}(x+3);}$$

so the partial fractions have the form

$${{2x^{2}+7}\over{(x-2)^{2}(x+3)}}={A\over{x-2}}+{B\over{(x-2)^{2}}}+{C\over{x+3}}={{A(x-2)(x+3)+B(x+3)+C(x-2)^{2}}\over{(x-2)^{2}(x+3)}}$$

and hence

$$A(x^{2}+x-6)+B(x+3)+C(x^{2}-4x+4)=2x^{2}+7.$$

We can find two of these values by setting $x=2$ and $x=-3$: in the former case we obtain $5B=15$ so $B=3$, and in the latter case we obtain $25C=25$ so $C=1$. To obtain the value of $A$ we inspect the constant term: $-6A+3B+4C=7$ so $-6A=-6$ and hence $A=1$.

Therefore

$${{{2x^{2}+7}\over{x^{3}-x^{2}-8x+12}}={{1}\over{x-2}}+{{3}\over{(x-2)^{2}}}+{{1}\over{x+3}}}$$

and the reqired integral is

$$\int{{2x^{2}+7}\over{x^{3}-x^{2}-8x+12}}\,dx=\int{{1}\over{x-2}}dx+\int{{3}\over{(x-2)^{2}}}dx+\int{{1}\over{x+3}}dx$$

$$=\log|x-2|-{{3}\over{x-2}}+\log|x+3|+K,$$

for some constant $K$.

ii) In this exercise we have to carry out all of the steps (a)-(e) in slide 1.25.

a) We first use division of polynomials to express the rational function ${{x^{3}-2}\over{x^{3}-x^{2}+4x-4}}$ as a sum of a polynomial and a rational function $\frac{f(x)}{g(x)}$ with $\deg f<\deg g$. In this case this step is rather easy, we have:

$${{x^{3}-2}\over{x^{3}-x^{2}+4x-4}}={{x^{3}-x^{2}+4x-4+(x^{2}-4x+2)}\over{x^{3}-x^{2}+4x-4}}=1+\frac{x^{2}-4x+2}{x^{3}-x^{2}+4x-4}.$$

b) Now we factorize the denominator $g(x)=x^{3}-x^{2}+4x-4$. We spot that $1$ is a root, so $(x-1)$ divides $g(x)$; then $g(x)=(x-1)(x^{2}+4)$.

c) To proceed, we have to express $\frac{x^{2}-4x+2}{(x-1)(x^{2}+4)}$ in the form $\frac{A}{x-1}+\frac{Bx+C}{x^{2}+4}$. Then we have

$$x^{2}-4x+2=A(x^{2}+4)+(Bx+C)(x-1)\;\Rightarrow(A+B)x^{2}+(C-B)x+(4A-C)=x^{2}-4x+2.$$

We can obtain the value of $A$ by setting $x=1$, but this method is not helpful for (directly) obtaining the values of $B$ and $C$. Setting $x=1$ we have $5A=-1$ and so $A=\frac{-1}{5}$. Now $A+B=1$, so $B=\frac{6}{5}$. Finally, $C-B=-4$ and so $C=B-4=\frac{-14}{5}$. (It is a good idea to check the linear term: we have $4A-C=\frac{-4}{5}+\frac{14}{5}=2$ as required.) Thus $\int{{x^{3}-2}\over{x^{3}-x^{2}+4x-4}}dx=\int dx-\frac{1}{5}\int\frac{dx}{x-1}+\int\frac{\frac{6}{5}x-\frac{14}{5}}{x^{2}+4}dx$.

d) There is one linear factor $(x-1)$, and the integral $\frac{-1}{5}\int\frac{dx}{x-1}=\frac{-1}{5}\log|x-1|+c$.

e) The general method for dealing with integrals such as $\int\frac{6x-14}{x^{2}+4}dx$ is outlined in slides 1.16-17 and demonstrated in slides 1.21-24. We first account for the linear term ($6x$) in the numerator by considering the substitution $u=x^{2}+4$, whence $\frac{du}{dx}=2x$ and so

$$\int\frac{6x\,dx}{x^{2}+4}=3\int\frac{du}{u}=3\log|u|+c^{\prime}=3\log(x^{2}+4)+c^{\prime}.$$

For the term $\frac{-14}{x^{2}+4}$ we make the substitution $x=2\tan t$ (or we remember the standard answer as above) to obtain $\int\frac{-14\,dx}{x^{2}+4}=-7\tan^{-1}\frac{x}{2}+c^{\prime\prime}$. Putting all of the different bits together, we obtain:

$$\int{{x^{3}-2}\over{x^{3}+3x^{2}+7x+5}}dx=x-\frac{1}{5}\log|x-1|+\frac{3}{5}\log(x^{2}+4)-\frac{7}{5}\tan^{-1}\left(\frac{x}{2}\right)+c.$$

W1.6 i) We observe that the coefficients of the the polynomial are real, so $1-3{\rm i}$ is also a root by Lemma 1.5 in the notes; hence

$$(z-1+3{\rm i})(z-1-3{\rm i})=(z-1)^{2}+9=z^{2}-2z+10$$

is a factor. We write

$$z^{4}-6z^{3}+26z^{2}-56z+80=(z^{2}-2z+10)(z^{2}+az+b)$$

where $a,b\in{\mathbb{R}}$. Since the constant term $10b=80$, we must have $b=8$. Now the coefficient of $z$ is $10a-2b=10a-16=-56$, so $a=-4$. Now $z^{2}-4z+8=0$ has roots

$$z={{4\pm\sqrt{16-32}}\over{2}}=2\pm\sqrt{-4};$$

so a complete list of the roots is

$$1+3{\rm i},1-3{\rm i},2+2{\rm i},2-2{\rm i}.$$

ii) Since the roots of the polynomials $x^{2}-2x+10$ and $x^{2}-4x+8$ are not real, they are both irreducible real polynomials, so the required factorization is $f(x)=(x^{2}-2x+10)(x^{2}-4x+8)$.

iii) We want to write $\frac{26}{f(x)}$ as

$$\frac{26}{f(x)}=\frac{\alpha x+\beta}{x^{2}-2x+10}+\frac{\gamma x+\delta}{x^{2}-4x+8}.$$

Multiplying through by $f(x)$, we obtain:

$$26=(\alpha x+\beta)(x^{2}-4x+8)+(\gamma x+\delta)(x^{2}-2x+10)$$

$$=(\alpha+\gamma)x^{3}+(\beta+\delta-4\alpha-2\gamma)x^{2}+(8\alpha+10\gamma-4\beta-2\delta)x+(8\beta+10\delta).$$

Thus $\gamma=-\alpha$ by looking at the coefficient of $x^{3}$. Substituting $-\alpha$ for $\gamma$, the coefficient of $x^{2}$ becomes $\beta+\delta-2\alpha=0$, so $\delta=2\alpha-\beta$. Now we substitute into the coefficient of $x$ and obtain $8\alpha-10\alpha-4\beta-4\alpha+2\beta=0$, so $-6\alpha-2\beta=0$. Therefore $\beta=-3\alpha$ and $\delta=5\alpha$. Now $8\beta+10\delta=-24\alpha+50\alpha=26\alpha$, so $\alpha=1$. Thus the required expression is

$$\frac{x-3}{x^{2}-2x+10}+\frac{5-x}{x^{2}-4x+8}.$$

iv) Using the partial fractions from the previous part, we have

$$\int\!\frac{26}{f(x)}dx=\int\!\frac{x-3}{x^{2}-2x+10}dx+\int\!\frac{5-x}{x^{2}-4x+8}dx.$$

For the first of these integrals, $Q(x)=x^{2}-2x+10$ and so we substitute $s=x-1$ to get $\frac{x-3}{x^{2}-2x+10}=\frac{s-2}{s^{2}+9}$. Clearly $\frac{dx}{ds}=1$ and so the first integral becomes

$$\int\frac{s-2}{s^{2}+9}\,ds=\frac{1}{2}\int\frac{s\,ds}{s^{2}+9}-\frac{2\,ds}{s^{2}+9}=\frac{1}{2}\log(s^{2}+9)-\frac{2}{3}\tan^{-1}\frac{s}{3}+c$$

where the last equality follows by making two separate substitutions ($u=s^{2}+9$ and $s=3\tan t$ respectively). We need to express this in terms of $x$: hence we have $\int\frac{x-3}{x^{2}-2x+10}\,dx=\frac{1}{2}\log(x^{2}-2x+10)-\frac{2}{3}\tan^{-1}\left(\frac{x-1}{3}\right)+c$.

For the second integral, we have $Q(x)=x^{2}-4x+8$ and so we first make the substitution $s=x-2$. Then $Q(x)=s^{2}+4$ and so $\frac{5-x}{x^{2}-4x+8}=\frac{3-s}{s^{2}+4}$. Hence we have

$$\int\frac{5-x}{x^{2}-4x+8}\,dx=\int\frac{3\,ds}{s^{2}+4}-\frac{s\,ds}{s^{2}+4}=\frac{3}{2}\tan^{-1}\frac{s}{2}-\frac{1}{2}\log(s^{2}+4)+c^{\prime}$$

Expressing this in terms of $x$ and putting everything together, we have

$$\int\!\frac{26}{f(x)}\,dx=\frac{1}{2}\log(x^{2}-2x+10)-\frac{2}{3}\tan^{-1}\left(\frac{x-1}{3}\right)-\frac{1}{2}\log(x^{2}-4x+8)+\frac{3}{2}\tan^{-1}\left(\frac{x-2}{2}\right)+c$$

$$=\frac{1}{2}\log\left(\frac{x^{2}-2x+10}{x^{2}-4x+8}\right)+\frac{3}{2}\tan^{-1}\left(\frac{x-2}{2}\right)-\frac{2}{3}\tan^{-1}\left(\frac{x-1}{3}\right)+c.$$