图中显示了 $x\ge 0$ 时的曲线 $y=\dfrac{1}{\sqrt{4x^2+4x+3}}$，以及一组由 $n$ 个宽为 $1$ 的矩形组成的图形。通过考虑这些矩形的面积之和，证明

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

题图是递减曲线下方的右端点矩形，宽为 $1$。把它们的面积和与积分比较，再配方，用反双曲正弦求积分。

Let $f(x)=(4x^2+4x+3)^{-1/2}$. The denominator is increasing for $x\ge0$, so $f$ is strictly decreasing. The right-endpoint rectangles on $[0,n]$ therefore give

The rectangle heights are $1/\sqrt{11}$, $1/\sqrt{27}$, $1/\sqrt{51}$, $\ldots$, $1/\sqrt{4(n-1)^2+4(n-1)+3}$ and $1/\sqrt{4n^2+4n+3}$. Completing the square gives $4x^2+4x+3=(2x+1)^2+2$. Put $u=(2x+1)/\sqrt2$, so $\mathrm{d}x=\mathrm{d}u/\sqrt2$ and $\sqrt{(2x+1)^2+2}=\sqrt2\sqrt{u^2+1}$. Thus the integral becomes $\frac12\int(u^2+1)^{-1/2}\,\mathrm{d}u$, with antiderivative $\frac12\operatorname{arsinh}u$. Hence

This gives the required bound.