CalcGospel 國際數學圖譜
返回

IAL 2025 Jan Q5

A Level / Edexcel / P4

IAL 2025 Jan Paper · Question 5

题目

Problem

(i) Find

∫x2e4x dx\int x^2e^{4x}\,dx

writing the answer in simplest form.

(4)

(ii) Use partial fractions and algebraic integration to show that

∫472x+11(2x+1)(2−x) dx=ln⁡k\int_4^7 \frac{2x+11}{(2x+1)(2-x)}\,dx=\ln k

where kk is a fully simplified rational constant to be found.

(6)
题目中文翻译

(i) 求

∫x2e4x dx\int x^2e^{4x}\,dx

并把答案写成最简形式。

(ii) 使用部分分式与代数积分，证明

∫472x+11(2x+1)(2−x) dx=ln⁡k\int_4^7 \frac{2x+11}{(2x+1)(2-x)}\,dx=\ln k

其中 kk 是一个需要求出的最简有理常数。

解答

(i)

We need to find

∫x2e4x dx\int x^2e^{4x}\,\mathrm{d}x

Use integration by parts, with

u=x2,dvdx=e4xu=x^2,\qquad \frac{dv}{dx}=e^{4x}

Then

dudx=2x,v=14e4x\frac{du}{dx}=2x,\qquad v=\frac14e^{4x}

So

∫x2e4x dx=14x2e4x−12∫xe4x dx\int x^2e^{4x}\,\mathrm{d}x =\frac14x^2e^{4x}-\frac12\int xe^{4x}\,\mathrm{d}x

Now use integration by parts again for

∫xe4x dx\int xe^{4x}\,\mathrm{d}x

Let

u=x,dvdx=e4xu=x,\qquad \frac{dv}{dx}=e^{4x}

Then

dudx=1,v=14e4x\frac{du}{dx}=1,\qquad v=\frac14e^{4x}

Therefore

∫xe4x dx=14xe4x−14∫e4x dx\int xe^{4x}\,\mathrm{d}x =\frac14xe^{4x}-\frac14\int e^{4x}\,\mathrm{d}x

So

∫xe4x dx=14xe4x−116e4x\int xe^{4x}\,\mathrm{d}x =\frac14xe^{4x}-\frac1{16}e^{4x}

Substitute this back:

∫x2e4x dx=14x2e4x−12(14xe4x−116e4x)=14x2e4x−18xe4x+132e4x+C\begin{aligned} \int x^2e^{4x}\,\mathrm{d}x &=\frac14x^2e^{4x} -\frac12\left(\frac14xe^{4x}-\frac1{16}e^{4x}\right) \\ &=\frac14x^2e^{4x}-\frac18xe^{4x}+\frac1{32}e^{4x}+C \end{aligned}

Hence

∫x2e4x dx=e4x(14x2−18x+132)+C\boxed{\int x^2e^{4x}\,\mathrm{d}x =e^{4x}\left(\frac14x^2-\frac18x+\frac1{32}\right)+C}

(ii)

First decompose into partial fractions:

2x+11(2x+1)(2−x)=A2x+1+B2−x\frac{2x+11}{(2x+1)(2-x)} =\frac{A}{2x+1}+\frac{B}{2-x}

Then

2x+11=A(2−x)+B(2x+1)2x+11=A(2-x)+B(2x+1)

Let x=2x=2:

15=5B15=5B

so

B=3B=3

Let x=−12x=-\dfrac12:

10=52A10=\frac52A

so

A=4A=4

Therefore

2x+11(2x+1)(2−x)=42x+1+32−x\frac{2x+11}{(2x+1)(2-x)} =\frac{4}{2x+1}+\frac{3}{2-x}

Now integrate:

∫(42x+1+32−x) dx=2ln⁡∣2x+1∣−3ln⁡∣2−x∣\int\left(\frac{4}{2x+1}+\frac{3}{2-x}\right)\,\mathrm{d}x =2\ln|2x+1|-3\ln|2-x|

So

∫472x+11(2x+1)(2−x) dx=[2ln⁡∣2x+1∣−3ln⁡∣2−x∣]47=(2ln⁡15−3ln⁡5)−(2ln⁡9−3ln⁡2)\begin{aligned} \int_4^7\frac{2x+11}{(2x+1)(2-x)}\,\mathrm{d}x &=\left[2\ln|2x+1|-3\ln|2-x|\right]_4^7 \\ &=(2\ln15-3\ln5)-(2\ln9-3\ln2) \end{aligned}

Therefore

∫472x+11(2x+1)(2−x) dx=2ln⁡15−3ln⁡5−2ln⁡9+3ln⁡2=ln⁡(152)−ln⁡(53)−ln⁡(92)+ln⁡(23)\begin{aligned} \int_4^7\frac{2x+11}{(2x+1)(2-x)}\,\mathrm{d}x &=2\ln15-3\ln5-2\ln9+3\ln2 \\ &=\ln(15^2)-\ln(5^3)-\ln(9^2)+\ln(2^3) \end{aligned}

Using logarithm laws,

∫472x+11(2x+1)(2−x) dx=ln⁡(152⋅2353⋅92)=ln⁡(180010125)=ln⁡(845)\begin{aligned} \int_4^7\frac{2x+11}{(2x+1)(2-x)}\,\mathrm{d}x &=\ln\left(\frac{15^2\cdot2^3}{5^3\cdot9^2}\right) \\ &=\ln\left(\frac{1800}{10125}\right) \\ &=\ln\left(\frac{8}{45}\right) \end{aligned}

Hence

k=845\boxed{k=\frac{8}{45}}