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IAL 2021 Jan FP3 Q6

A Level / Edexcel / FP3

IAL 2021 Jan Paper · Question 6

题目

Problem

Let

In=∫xnx2+3 dx,n∈N.I_n=\int\frac{x^n}{\sqrt{x^2+3}}\,\mathrm{d}x, \qquad n\in\mathbb{N}.

(a) Show that

In=xn−1nx2+3−3(n−1)nIn−2,n≥3.I_n=\frac{x^{n-1}}{n}\sqrt{x^2+3} -\frac{3(n-1)}{n}I_{n-2}, \qquad n\ge3.

(b) Hence show that

∫x5x2+3 dx=15x2+3 (x4+px2+q)+k,\int\frac{x^5}{\sqrt{x^2+3}}\,\mathrm{d}x =\frac15\sqrt{x^2+3}\,(x^4+px^2+q)+k,

where pp and qq are integers to be determined and kk is an arbitrary constant.

(10)
题目中文翻译

设

In=∫xnx2+3 dx,n∈N.I_n=\int\frac{x^n}{\sqrt{x^2+3}}\,\mathrm{d}x, \qquad n\in\mathbb{N}.

(a) 证明

In=xn−1nx2+3−3(n−1)nIn−2,n≥3.I_n=\frac{x^{n-1}}{n}\sqrt{x^2+3} -\frac{3(n-1)}{n}I_{n-2}, \qquad n\ge3.

(b) 因此证明

∫x5x2+3 dx=15x2+3 (x4+px2+q)+k,\int\frac{x^5}{\sqrt{x^2+3}}\,\mathrm{d}x =\frac15\sqrt{x^2+3}\,(x^4+px^2+q)+k,

其中 pp 和 qq 为待定整数，kk 为任意常数。

解答

(a)

解法一

思路

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把 xnx^n 写成 xn−1⋅xx^{n-1}\cdot x，再分部积分。这样 x/x2+3x/\sqrt{x^2+3} 可以直接积分，而余下积分中的 x2+3\sqrt{x^2+3} 可改写为 (x2+3)/x2+3(x^2+3)/\sqrt{x^2+3}，从而同时出现 InI_n 和 In−2I_{n-2}。

答题过程

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Write

In=∫xn−1xx2+3 dx.I_n=\int x^{n-1} \frac{x}{\sqrt{x^2+3}}\,\mathrm{d}x.

Integrating by parts with

u=xn−1,dv=xx2+3 dx,u=x^{n-1}, \qquad \mathrm{d}v=\frac{x}{\sqrt{x^2+3}}\,\mathrm{d}x,

gives

du=(n−1)xn−2 dx,v=x2+3.\mathrm{d}u=(n-1)x^{n-2}\,\mathrm{d}x, \qquad v=\sqrt{x^2+3}.

Therefore

In= xn−1x2+3 −(n−1)∫xn−2x2+3 dx.\begin{align*} I_n =&\,x^{n-1}\sqrt{x^2+3}\\[4mm] &\,\hspace{2pt}-(n-1) \int x^{n-2}\sqrt{x^2+3}\,\mathrm{d}x. \end{align*}

Now

 ∫xn−2x2+3 dx= ∫xn−2(x2+3)x2+3 dx= In+3In−2.\begin{align*} &\,\int x^{n-2}\sqrt{x^2+3}\,\mathrm{d}x\\[4mm] =&\,\int \frac{x^{n-2}(x^2+3)}{\sqrt{x^2+3}}\,\mathrm{d}x\\[4mm] =&\,I_n+3I_{n-2}. \end{align*}

Hence

In=xn−1x2+3−(n−1)In−3(n−1)In−2.I_n=x^{n-1}\sqrt{x^2+3} -(n-1)I_n-3(n-1)I_{n-2}.

Thus

nIn=xn−1x2+3−3(n−1)In−2,nI_n=x^{n-1}\sqrt{x^2+3} -3(n-1)I_{n-2},

and so

In=xn−1nx2+3−3(n−1)nIn−2,n≥3.\boxed{I_n=\frac{x^{n-1}}{n}\sqrt{x^2+3} -\frac{3(n-1)}{n}I_{n-2}}, \qquad n\ge3.

解法二

思路

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另一条官方路线先写 xn=xn−2(x2+3−3)x^n=x^{n-2}(x^2+3-3)，将 InI_n 拆成一个含 x2+3\sqrt{x^2+3} 的积分和 In−2I_{n-2}。再对前者反向选择分部积分，最终同样整理出递推式。

答题过程

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Since x2=(x2+3)−3x^2=(x^2+3)-3,

In= ∫xn−2x2x2+3 dx= ∫xn−2x2+3 dx−3In−2.\begin{align*} I_n =&\,\int \frac{x^{n-2}x^2}{\sqrt{x^2+3}}\,\mathrm{d}x\\[4mm] =&\,\int x^{n-2}\sqrt{x^2+3}\,\mathrm{d}x -3I_{n-2}. \end{align*}

Let

J=∫xn−2x2+3 dx.J=\int x^{n-2}\sqrt{x^2+3}\,\mathrm{d}x.

Integrating by parts with

u=x2+3,dv=xn−2 dx,u=\sqrt{x^2+3}, \qquad \mathrm{d}v=x^{n-2}\,\mathrm{d}x,

gives

J= xn−1n−1x2+3 −1n−1∫xnx2+3 dx= xn−1n−1x2+3−1n−1In.\begin{align*} J =&\,\frac{x^{n-1}}{n-1}\sqrt{x^2+3}\\[4mm] &\,\hspace{2pt}-\frac1{n-1} \int\frac{x^n}{\sqrt{x^2+3}}\,\mathrm{d}x\\[4mm] =&\,\frac{x^{n-1}}{n-1}\sqrt{x^2+3} -\frac1{n-1}I_n. \end{align*}

Substituting this into In=J−3In−2I_n=J-3I_{n-2},

In=xn−1n−1x2+3−1n−1In−3In−2.I_n=\frac{x^{n-1}}{n-1}\sqrt{x^2+3} -\frac1{n-1}I_n-3I_{n-2}.

Multiplying by n−1n-1 and rearranging gives

In=xn−1nx2+3−3(n−1)nIn−2.\boxed{I_n=\frac{x^{n-1}}{n}\sqrt{x^2+3} -\frac{3(n-1)}{n}I_{n-2}}.

(b)

解法一

思路

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依次在递推公式中取 n=5n=5 和 n=3n=3，把 I5I_5 降到 I3I_3，再降到容易直接积分的 I1I_1。这直接体现题目中 Hence 的承接关系。

答题过程

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Using the reduction formula with n=5n=5,

I5=x45x2+3−125I3.I_5=\frac{x^4}{5}\sqrt{x^2+3} -\frac{12}{5}I_3.

With n=3n=3,

I3=x23x2+3−2I1.I_3=\frac{x^2}{3}\sqrt{x^2+3}-2I_1.

Also,

I1=∫xx2+3 dx=x2+3.I_1=\int\frac{x}{\sqrt{x^2+3}}\,\mathrm{d}x =\sqrt{x^2+3}.

Hence

I3= x23x2+3−2x2+3= 13(x2−6)x2+3.\begin{align*} I_3 =&\,\frac{x^2}{3}\sqrt{x^2+3} -2\sqrt{x^2+3}\\[4mm] =&\,\frac13(x^2-6)\sqrt{x^2+3}. \end{align*}

Substituting into the expression for I5I_5,

I5= x45x2+3−45(x2−6)x2+3= 15x2+3(x4−4x2+24)+k.\begin{align*} I_5 =&\,\frac{x^4}{5}\sqrt{x^2+3} -\frac45(x^2-6)\sqrt{x^2+3}\\[4mm] =&\,\frac15\sqrt{x^2+3} \bigl(x^4-4x^2+24\bigr)+k. \end{align*}

Therefore

p=−4,q=24.\boxed{p=-4,\qquad q=24}.

解法二

思路

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官方也允许只先用一次递推公式，把 I5I_5 化到 I3I_3，再以 u=x2+3u=x^2+3 直接求 I3I_3。这仍然使用了 (a)，所以符合 Hence 的要求。

答题过程

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From the reduction formula,

I5=x45x2+3−125I3.I_5=\frac{x^4}{5}\sqrt{x^2+3} -\frac{12}{5}I_3.

To evaluate I3I_3, let u=x2+3u=x^2+3. Then du=2x dx\mathrm{d}u=2x\,\mathrm{d}x and x2=u−3x^2=u-3, so

I3= ∫x3x2+3 dx= 12∫(u−3)u−1/2 du= 12∫(u1/2−3u−1/2) du= 13u3/2−3u1/2= 13(x2−6)x2+3.\begin{align*} I_3 =&\,\int\frac{x^3}{\sqrt{x^2+3}}\,\mathrm{d}x\\[4mm] =&\,\frac12\int(u-3)u^{-1/2}\,\mathrm{d}u\\[4mm] =&\,\frac12\int \bigl(u^{1/2}-3u^{-1/2}\bigr)\,\mathrm{d}u\\[4mm] =&\,\frac13u^{3/2}-3u^{1/2}\\[4mm] =&\,\frac13(x^2-6)\sqrt{x^2+3}. \end{align*}

Therefore

I5= x45x2+3−45(x2−6)x2+3= 15x2+3(x4−4x2+24)+k.\begin{align*} I_5 =&\,\frac{x^4}{5}\sqrt{x^2+3} -\frac45(x^2-6)\sqrt{x^2+3}\\[4mm] =&\,\frac15\sqrt{x^2+3} \bigl(x^4-4x^2+24\bigr)+k. \end{align*}

Thus

p=−4,q=24.\boxed{p=-4,\qquad q=24}.