$\int \frac{d t}{\left(t^{2}+\frac{3}{4}\right)^{2}}\left(\text { Put } x+\frac{1}{2}=t\right)$
$=\frac{\sqrt{3}}{2} \int \frac{\sec ^{2} \theta d \theta}{\frac{9}{16} \sec ^{4} \theta}\left(\operatorname{Put} t=\frac{\sqrt{3}}{2} \tan \theta\right)$
$=\frac{4 \sqrt{3}}{9} \int(1+\cos 2 \theta) d \theta$
$=\frac{4 \sqrt{3}}{9}\left[\theta+\frac{\sin 2 \theta}{2}\right]+\mathrm{c}$
$=\frac{4 \sqrt{3}}{9}\left[\tan ^{-1}\left(\frac{2 \mathrm{x}+1}{\sqrt{3}}\right)+\frac{\sqrt{3}(2 \mathrm{x}+1)}{3+(2 \mathrm{x}+1)^{2}}\right]+\mathrm{c}$
$=\frac{4 \sqrt{3}}{9} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+\frac{1}{3}\left(\frac{2 x+1}{x^{2}+x+1}\right)+c$
Hence, $9(\sqrt{3} \mathrm{a}+\mathrm{b})=15$
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