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IAL 2021 Jan Q9

A Level / Edexcel / P3

IAL 2021 Jan Paper · Question 9

题目

Problem

Find

(i)

∫3x−23x2−4x+5 dx\int \frac{3x-2}{3x^2-4x+5}\,dx
(2)

(ii)

∫e2x(e2x−1)3 dxx≠0\int \frac{e^{2x}}{(e^{2x}-1)^3}\,dx \qquad x\ne 0
(2)
题目中文翻译

求

(i)

∫3x−23x2−4x+5 dx\int \frac{3x-2}{3x^2-4x+5}\,dx

(ii)

∫e2x(e2x−1)3 dxx≠0\int \frac{e^{2x}}{(e^{2x}-1)^3}\,dx \qquad x\ne 0

解答

(i)

We need to find

∫3x−23x2−4x+5 dx\int \frac{3x-2}{3x^2-4x+5}\,\mathrm{d}x

Notice that the derivative of the denominator is

ddx(3x2−4x+5)=6x−4\frac{d}{dx}(3x^2-4x+5)=6x-4

and

6x−4=2(3x−2)6x-4=2(3x-2)

So the numerator is half the derivative of the denominator.

Therefore

∫3x−23x2−4x+5 dx=12∫6x−43x2−4x+5 dx=12ln⁡(3x2−4x+5)+c\begin{aligned} \int \frac{3x-2}{3x^2-4x+5}\,\mathrm{d}x &=\frac12\int \frac{6x-4}{3x^2-4x+5}\,\mathrm{d}x\\ &=\frac12\ln(3x^2-4x+5)+c \end{aligned}

Hence

12ln⁡(3x2−4x+5)+c\boxed{\frac12\ln(3x^2-4x+5)+c}

(ii)

We need to find

∫e2x(e2x−1)3 dx\int \frac{e^{2x}}{(e^{2x}-1)^3}\,\mathrm{d}x

Let

u=e2x−1u=e^{2x}-1

Then

dudx=2e2x\frac{du}{dx}=2e^{2x}

so

e2x dx=12 due^{2x}\,\mathrm{d}x=\frac12\,\mathrm{d}u

Therefore

∫e2x(e2x−1)3 dx=12∫u−3 du=12⋅u−2−2=−14u−2\begin{aligned} \int \frac{e^{2x}}{(e^{2x}-1)^3}\,\mathrm{d}x &=\frac12\int u^{-3}\,\mathrm{d}u\\ &=\frac12\cdot \frac{u^{-2}}{-2}\\ &=-\frac14u^{-2} \end{aligned}

Substitute back:

−14(e2x−1)2+c\boxed{-\frac{1}{4(e^{2x}-1)^2}+c}