10.1 Solutions to Assessed Exercises 1

A1.1 i) Clearly $g(x)$ is divisible by $x$. Furthermore we have $g(-2)=16-16+16-16=0$, so $g(x)$ is also divisible by $(x+2)$. Dividing through, we find that $g(x)=x(x+2)(x^{2}+4)$. Clearly $x^{2}+4$ is irreducible, so we have obtained the required factorization. [1]

ii) Using our answer to (i), we have to express $\frac{x^{3}+5x^{2}-6x+8}{x^{4}+2x^{3}+4x^{2}+8x}$ in the form

$$\frac{A}{x}+\frac{B}{x+2}+\frac{Cx+D}{x^{2}+4}.$$

Multiplying through by $g(x)$, we have

$$A(x+2)(x^{2}+4)+Bx(x^{2}+4)+(Cx+D)x(x+2)=x^{3}+5x^{2}-6x+8.\;\;\;\mbox{(*)}$$

We can use the trick of setting $x=0$ to obtain the value of $A$, and we can set $x=-2$ to obtain the value of $B$. (This trick will not help us to find the values of $C$ and $D$.) Setting $x=0$, we have $8A=8$ and therefore $A=1$. Setting $x=-2$, we obtain $-16B=-8+20+12+8=32$, so $B=-2$. To obtain $C$ and $D$ we can either equate the coefficients of $x$ and $x^{2}$, or we can subtract $A(x+2)(x^{2}+4)+Bx(x^{2}+4)$ from both sides of (*). Using the latter argument, we see that

$$(Cx+D)x(x+2)=x^{3}+5x^{2}-6x+8-(x+2)(x^{2}+4)+2x(x^{2}+4)=2x^{3}+3x^{2}-2x=(2x-1)x(x+2).$$

Thus $Cx+D=x(x+2)$. (If equating coefficients, then we see that the coefficient of $x$ on the left-hand side of (*) is $4A+4B+2D=2D-4$, so that $D=-1$, and the coefficient of $x^{2}$ on the left-hand side of (*) is $2A+2C+D=2C+1$, so that $C=2$.)

(Total [3] marks for the partial fractions step: 2 marks if mostly correct with one or two minor errors; at most 1 mark if the $Cx$ term is omitted from the partial fractions expression to be obtained.)

Now we integrate:

$$\int_{2\sqrt{3}/3}^{2\sqrt{3}}\frac{x^{3}+5x^{2}-6x+8}{x^{4}+2x^{3}+4x^{2}+8x}\,dx=\int_{2\sqrt{3}/3}^{2\sqrt{3}}\left(\frac{1}{x}-\frac{2}{x+2}+\frac{2x-1}{x^{2}+4}\right)\,dx.$$

The integral of the first term is $\left[\log|x|\right]_{2\sqrt{3}/3}^{2\sqrt{3}}=\log\left(\frac{2\sqrt{3}}{2\sqrt{3}/3}\right)=\log 3$. The integral of the second term is $-2\left[\log|x+2|\right]_{2\sqrt{3}/3}^{2\sqrt{3}}=-2\log\frac{2\sqrt{3}+2}{2\sqrt{3}/3+2}=-2\log\sqrt{3}=-\log 3$.

For the third term, we need to make two different substitutions to integrate the $\frac{2x}{x^{2}+4}$ and $\frac{-1}{x^{2}+4}$ terms separately. Substituting $u=x^{2}+4$ for the first of these, we obtain $\int_{2\sqrt{3}/3}^{2\sqrt{3}}\frac{2x\,dx}{x^{2}+4}=\left[\log(x^{2}+4)\right]_{2\sqrt{3}/3}^{2\sqrt{3}}=\log\frac{16}{16/3}=\log 3$. Finally, substituting $x=2\tan t$, we have

$$\int_{2\sqrt{3}/3}^{2\sqrt{3}}\frac{-dx}{x^{2}+4}=-\frac{1}{2}\left[\tan^{-1}\frac{x}{2}\right]_{2\sqrt{3}/3}^{2\sqrt{3}}=-\frac{1}{2}\left(\tan^{-1}\sqrt{3}-\tan^{-1}\frac{\sqrt{3}}{2}\right)=-\frac{1}{2}\left(\frac{\pi}{3}-\frac{\pi}{6}\right)=-\frac{\pi}{12}.$$

Summing the terms, the value of the integral is $\log 3-\frac{\pi}{12}$.

(Total [3] marks for the integration step: $\frac{1}{2}$ for the integral of each of the terms $\frac{1}{x}$, $\frac{-2}{x+2}$, $\frac{2x}{x^{2}+4}$, $\frac{-1}{x^{2}+4}$ and $1$ for the sum.)

A1.2 We have $x=a\sinh u$, so ${{dx}\over{du}}=a\cosh u$ and $a^{2}+x^{2}=a^{2}(1+\sinh^{2}u)=a^{2}\cosh^{2}u$; let $t=a\sinh v$ for simplicity. Then, using the identities in the question, we have

$$\int_{0}^{t}(a^{2}+x^{2})^{1/2}dx=\int_{0}^{v}a^{2}\cosh^{2}u\,du={{a^{2}}\over{2}}\int_{0}^{v}(1+\cosh 2u)\,du$$

$$={{a^{2}}\over{2}}\Bigl[u+{{1}\over{2}}\sinh 2u\Bigr]_{0}^{v}={{a^{2}}\over{2}}\Bigl(v+{{1}\over{2}}\sinh 2v\Bigr)={{a^{2}}\over{2}}\Bigl(v+\sinh v\cosh v\Bigr)$$

$$={{a^{2}}\over{2}}\Bigl(v+{{t}\over{a}}\Bigl(1+{{t^{2}}\over{a^{2}}}\Bigr)^{1/2}\Bigr)={{a^{2}v}\over{2}}+{{1}\over{2}}t\sqrt{a^{2}+t^{2}}={{a^{2}}\over{2}}\sinh^{-1}{{t}\over{a}}+{{1}\over{2}}t\sqrt{a^{2}+t^{2}}.$$

(Total [3] marks: 1 for making the substitution correctly, 1 for getting the indefinite integral right (i.e. obtaining $\frac{a^{2}}{2}(v+\frac{1}{2}\sinh 2v)$) and 1 for evaluating. Only lose half a mark for failing to simplify $\sinh(2\sinh^{-1}\frac{t}{a})$.)