(a) Resolve into partial fractions then integrate: $\ln|x+1|+\tfrac{2}{x+1}+3\ln|x-3|+C$. (b) Standardise heights; counts are $46,\ 46,\ 341$ out of 500.

Part (a) — integration by partial fractions. Write

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

Then $4x^{2}+2x+6=A(x+1)(x-3)+B(x-3)+C(x+1)^{2}$.

• $x=-1:\ 4-2+6=8=B(-4)\Rightarrow B=-2.$
• $x=3:\ 36+6+6=48=C(16)\Rightarrow C=3.$
• Comparing $x^{2}$: $4=A+C\Rightarrow A=1.$

So

$\int\frac{4x^{2}+2x+6}{(x+1)^{2}(x-3)}\,dx=\int\left[\frac{1}{x+1}-\frac{2}{(x+1)^{2}}+\frac{3}{x-3}\right]dx.$

$=\ln|x+1|-2\cdot\frac{(x+1)^{-1}}{-1}+3\ln|x-3|+C=\ln|x+1|+\frac{2}{x+1}+3\ln|x-3|+C.$

Part (b) — normal distribution. $N=500$, mean $\mu=68$, $\sigma=3$, $Z=\dfrac{X-68}{3}$. Given areas: area$(0\text{ to }1)=0.3413$, area$(0\text{ to }1.33)=0.4082$.

(i) Height $>72$: $Z=\dfrac{72-68}{3}=1.33$.

$P(Z>1.33)=0.5-0.4082=0.0918\Rightarrow 500\times0.0918=45.9\approx46\text{ students}.$

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