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IAL 2025 Jan FP3 Q3

A Level / Edexcel / FP3

IAL 2025 Jan Paper · Question 3

题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

(i) Determine

∫14x2+12x+25 dx\int \frac{1}{4x^2 + 12x + 25}\,dx

(ii) Show that

∫391x2+4x−17 dx=ln⁡a\int_3^9 \frac{1}{\sqrt{x^2 + 4x - 17}}\,dx = \ln a

where aa is an integer to be determined.

(4)
(6)
题目中文翻译

在本题中，你必须写出所有解题步骤。

完全依赖计算器技术求解是不可以的。

(i) 求

∫14x2+12x+25 dx\int \frac{1}{4x^2 + 12x + 25}\,dx

(ii) 证明

∫391x2+4x−17 dx=ln⁡a\int_3^9 \frac{1}{\sqrt{x^2 + 4x - 17}}\,dx = \ln a

其中 aa 为待确定的整数。

解答

(i)

解法一

思路

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先把二次式配成 (2x+3)2+42(2x+3)^2+4^2，再套用 ∫1u2+a2 du=1aarctan⁡ua+C\int\frac{1}{u^2+a^2}\,\mathrm du=\frac1a\arctan\frac ua+C。换元时要同时保留 du=2 dx\mathrm du=2\,\mathrm dx 带来的系数。

答题过程

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Completing the square,

4x2+12x+25= (2x+3)2+16.\begin{align*} 4x^2+12x+25 =&\,(2x+3)^2+16. \end{align*}

Let u=2x+3u=2x+3, so du=2 dx\mathrm du=2\,\mathrm dx. Then

∫14x2+12x+25 dx= 12∫1u2+42 du= 18arctan⁡(u4)+C.\begin{align*} \int\frac{1}{4x^2+12x+25}\,\mathrm dx =&\,\frac12\int\frac{1}{u^2+4^2}\,\mathrm du\\ =&\,\frac18\arctan\left(\frac u4\right)+C. \end{align*}

Therefore,

18arctan⁡(2x+34)+C.\boxed{ \frac18\arctan\left(\frac{2x+3}{4}\right)+C}.

(ii)

解法一

思路

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先配方得到 (x+2)2−21(x+2)^2-21，再使用 ∫1u2−a2 du\int\frac{1}{\sqrt{u^2-a^2}}\,\mathrm du 的对数形式。由于积分区间内 x+2>21x+2>\sqrt{21}，根式及对数参数均为正；代入上下限后必须继续合并成单一对数。

答题过程

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Completing the square,

x2+4x−17=(x+2)2−21.x^2+4x-17=(x+2)^2-21.

On 3⩽x⩽93\leqslant x\leqslant9, x+2>21x+2>\sqrt{21}. Hence

 ∫391x2+4x−17 dx= [ln⁡(x+2+(x+2)2−21)]39= ln⁡(11+121−21) −ln⁡(5+25−21)= ln⁡21−ln⁡7= ln⁡3.\begin{align*} &\,\int_3^9 \frac{1}{\sqrt{x^2+4x-17}}\,\mathrm dx\\ =&\,\left[ \ln\Big( x+2+\sqrt{(x+2)^2-21} \Big) \right]_3^9\\ =&\,\ln(11+\sqrt{121-21})\\ &\,\hspace{2pt}-\ln(5+\sqrt{25-21})\\ =&\,\ln21-\ln7\\ =&\,\ln3. \end{align*}

Therefore,

a=3.\boxed{a=3}.

解法二

思路

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官方评分资料也允许双曲代换。令 x+2=21cosh⁡ux+2=\sqrt{21}\cosh u 后，分母与 dx\mathrm dx 中的 21sinh⁡u\sqrt{21}\sinh u 会约去，积分化成 ∫du\int\mathrm du；最后用反双曲余弦的对数定义化简。

答题过程

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Let

x+2=21cosh⁡u.x+2=\sqrt{21}\cosh u.

Since x+2>21x+2>\sqrt{21} throughout the interval, take u>0u>0. Then

dx=21sinh⁡u du\mathrm dx=\sqrt{21}\sinh u\,\mathrm du

and

(x+2)2−21=21sinh⁡u.\sqrt{(x+2)^2-21} =\sqrt{21}\sinh u.

The limits become

u=arcosh⁡(521)u=\operatorname{arcosh}\left(\frac5{\sqrt{21}}\right)

and

u=arcosh⁡(1121).u=\operatorname{arcosh}\left(\frac{11}{\sqrt{21}}\right).

Therefore,

 ∫391x2+4x−17 dx= ∫arcosh⁡(5/21)arcosh⁡(11/21)1 du= arcosh⁡(1121) −arcosh⁡(521).\begin{align*} &\,\int_3^9 \frac{1}{\sqrt{x^2+4x-17}}\,\mathrm dx\\ =&\,\int_{ \operatorname{arcosh}(5/\sqrt{21})}^{ \operatorname{arcosh}(11/\sqrt{21})} 1\,\mathrm du\\ =&\,\operatorname{arcosh} \left(\frac{11}{\sqrt{21}}\right)\\ &\,\hspace{2pt} -\operatorname{arcosh} \left(\frac5{\sqrt{21}}\right). \end{align*}

Using

arcosh⁡z=ln⁡(z+z2−1),\operatorname{arcosh}z =\ln\left(z+\sqrt{z^2-1}\right),

this becomes

 ln⁡(11+1021)−ln⁡(5+221)= ln⁡(217)= ln⁡3.\begin{align*} &\,\ln\left(\frac{11+10}{\sqrt{21}}\right) -\ln\left(\frac{5+2}{\sqrt{21}}\right)\\ =&\,\ln\left(\frac{21}{7}\right)\\ =&\,\ln3. \end{align*}

Hence

a=3.\boxed{a=3}.