2022HSCShort answer

Q12 (d)

4 marks

Question

Using partial fractions, evaluate
$\int_2^n\frac{4+x}{(1-x)(4+x^2)}\,dx,$
giving your answer in the form
$\frac12\ln\left(\frac{f(n)}{8(n-1)^2}\right),$
where $f(n)$ is a function of $n$ .

Try the question before revealing the answer.

Solution

Criteria	Marks
Provides correct solution	4
Obtains correct primitive	3
Obtains the correct partial fraction decomposition, or equivalent merit	2
Writes the correct general form of the partial fraction decomposition, or equivalent merit	1

Sample answer:

\frac{4+x}{(1-x)(4+x^2)}=\frac{A}{1-x}+\frac{Bx+C}{4+x^2}

$4+x=A(4+x^2)+(Bx+C)(1-x)$
$x=1:\quad5=5A,\quad A=1$

\frac{4+x}{(1-x)(4+x^2)}-\frac1{1-x} =\frac{x-x^2}{(1-x)(4+x^2)}=\frac{x}{4+x^2}

\int_2^n\frac{4+x}{(1-x)(4+x^2)}dx =\int_2^n\left(\frac1{1-x}+\frac{x}{4+x^2}\right)dx

=\int_2^n\left(\frac1{1-x}+\frac12\frac{2x}{4+x^2}\right)dx

=\left[-\ln|1-x|+\frac12\ln|4+x^2|\right]_2^n

=-\ln|1-n|+\frac12\ln|4+n^2|-\frac12\ln8

=\ln\frac1{|n-1|}+\frac12\ln|4+n^2|-\frac12\ln8

=\frac12\ln\left(\frac{4+n^2}{8|n-1|^2}\right)

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Syllabus alignment

Calculus

Further IntegrationApply substitution, partial fractions, integration by parts and recurrence relationships to theoretical and practical problems.

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