题目

Problem

(i) Find

\int \frac{12}{(2x-1)^2}\,dx

giving your answer in simplest form.

(2)

(ii) (a) Write

\frac{4x+3}{x+2}

in the form

A+\frac{B}{x+2}

where $A$ and $B$ are constants to be found.

(b) Hence find, using algebraic integration, the exact value of

\int_{-8}^{-5}\frac{4x+3}{x+2}\,dx

giving your answer in simplest form.

(6)

题目中文翻译

(i) 求

\int \frac{12}{(2x-1)^2}\,dx

并将答案化为最简形式。

(ii) (a) 将

\frac{4x+3}{x+2}

写成

A+\frac{B}{x+2}

的形式，其中 $A$ 和 $B$ 为待求常数。

(b) 由此利用代数积分法求

\int_{-8}^{-5}\frac{4x+3}{x+2}\,dx

的精确值，并将答案化为最简形式。

解答

We need to find

\int \frac{12}{(2x-1)^2}\,\mathrm{d}x

Write the integrand using a negative power:

\frac{12}{(2x-1)^2}=12(2x-1)^{-2}

Since the derivative of $2x-1$ is $2$ ,

\begin{aligned} \int 12(2x-1)^{-2}\,\mathrm{d}x &=12\cdot \frac{(2x-1)^{-1}}{-1}\cdot \frac12\\ &=-6(2x-1)^{-1}\\ &=-\frac{6}{2x-1} \end{aligned}

Therefore

\boxed{\int \frac{12}{(2x-1)^2}\,\mathrm{d}x=-\frac{6}{2x-1}+c}

We want

\frac{4x+3}{x+2}=A+\frac{B}{x+2}

Multiply by $x+2$ :

4x+3=A(x+2)+B

Expand:

4x+3=Ax+2A+B

Compare coefficients:

A=4

and

2A+B=3

8+B=3

which gives

B=-5

Therefore

\boxed{\frac{4x+3}{x+2}=4-\frac{5}{x+2}}

Using part (ii)(a),

\int_{-8}^{-5}\frac{4x+3}{x+2}\,\mathrm{d}x =\int_{-8}^{-5}\left(4-\frac{5}{x+2}\right)\,\mathrm{d}x

Integrate:

\int \left(4-\frac{5}{x+2}\right)\,\mathrm{d}x =4x-5\ln|x+2|

Therefore

\begin{aligned} \int_{-8}^{-5}\frac{4x+3}{x+2}\,\mathrm{d}x &=\left[4x-5\ln|x+2|\right]_{-8}^{-5}\\ &=\left(-20-5\ln3\right)-\left(-32-5\ln6\right)\\ &=12+5\ln6-5\ln3\\ &=12+5\ln2 \end{aligned}

Hence

\boxed{12+5\ln2}