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CIE 9709 2025 November Paper 32 Q4

A Level / CIE / P3

CIE 9709 2025 Nov Paper 32 Paper · Question 4

题目

Problem

Find the exact value of ∫01xtan⁡−1x dx\displaystyle\int_0^1x\tan^{-1}x\,\mathrm{d}x.

(6)
题目中文翻译

求 ∫01xtan⁡−1x dx\displaystyle\int_0^1x\tan^{-1}x\,\mathrm{d}x 的精确值。

(6)
原卷通用作答要求

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。

解答

解法一

思路

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分部积分时对反正切求导、对 xx 积分。余下分式先拆成一与简单分式,再积分。

答题过程

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Let u=tan⁡−1xu=\tan^{-1}x and dv=x dx\mathrm{d}v=x\,\mathrm{d}x. Integration by parts gives

I= [x22tan⁡−1x]01 −12∫01x21+x2 dx= π8−12∫01(1−11+x2) dx= π8−12[x−tan⁡−1x]01= π8−12(1−π4).\begin{align*} I=&\,\bigg[\frac{x^2}{2}\tan^{-1}x\bigg]_0^1\\[4mm] &\,\hspace{2pt}-\frac12\int_0^1\frac{x^2}{1+x^2}\,\mathrm{d}x\\[4mm] =&\,\frac\pi8-\frac12\int_0^1 \bigg(1-\frac1{1+x^2}\bigg)\,\mathrm{d}x\\ =&\,\frac\pi8-\frac12\big[x-\tan^{-1}x\big]_0^1\\ =&\,\frac\pi8-\frac12\bigg(1-\frac\pi4\bigg). \end{align*}

The lower-limit contributions are zero because x=0x=0 and tan⁡−10=0\tan^{-1}0=0. Hence

I=π4−12.\boxed{I=\frac\pi4-\frac12}.