题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

f(x)=\frac{2x^3-4x-15}{x^2+3x+4}

(a) Show that

f(x)\equiv Ax+B+\frac{C(2x+3)}{x^2+3x+4}

where $A$ , $B$ and $C$ are integers to be found.

(4)

(b) Hence, find

\int_3^5 f(x)\,dx

giving your answer in the form $p+\ln q$ , where $p$ and $q$ are integers.

(5)

题目中文翻译

本题中你必须写出解题过程的所有步骤。

不接受完全依赖计算器技术的解法。

f(x)=\frac{2x^3-4x-15}{x^2+3x+4}

(a) 证明

f(x)\equiv Ax+B+\frac{C(2x+3)}{x^2+3x+4}

其中 $A$ 、 $B$ 和 $C$ 是需要求出的整数。

(b) 由此求

\int_3^5 f(x)\,dx

并把答案写成 $p+\ln q$ 的形式，其中 $p$ 和 $q$ 是整数。

解答

We want

\frac{2x^3-4x-15}{x^2+3x+4} \equiv Ax+B+\frac{C(2x+3)}{x^2+3x+4}

Multiply both sides by

x^2+3x+4

to get

2x^3-4x-15 =(Ax+B)(x^2+3x+4)+C(2x+3)

Expand:

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

2x^3-4x-15 =Ax^3+(3A+B)x^2+(4A+3B+2C)x+(4B+3C)

Compare coefficients.

For $x^3$ :

A=2

For $x^2$ :

3A+B=0

Since $A=2$ ,

6+B=0

B=-6

For the constant term:

4B+3C=-15

Substitute $B=-6$ :

-24+3C=-15

Thus

C=3

Therefore

\boxed{A=2,\qquad B=-6,\qquad C=3}

and

f(x)\equiv 2x-6+\frac{3(2x+3)}{x^2+3x+4}

Using part (a),

\int_3^5 f(x)\,\mathrm{d}x = \int_3^5\left(2x-6+\frac{3(2x+3)}{x^2+3x+4}\right)\,\mathrm{d}x

Now

\int\left(2x-6+\frac{3(2x+3)}{x^2+3x+4}\right)\,\mathrm{d}x =x^2-6x+3\ln(x^2+3x+4)

Therefore

\begin{aligned} \int_3^5 f(x)\,\mathrm{d}x &=\left[x^2-6x+3\ln(x^2+3x+4)\right]_3^5 \\ &=\left(25-30+3\ln44\right) -\left(9-18+3\ln22\right) \end{aligned}

\int_3^5 f(x)\,\mathrm{d}x =4+3\ln44-3\ln22

Using logarithm laws,

3\ln44-3\ln22 =3\ln\left(\frac{44}{22}\right)

Thus

3\ln\left(\frac{44}{22}\right) =3\ln2 =\ln8

Hence

\boxed{\int_3^5 f(x)\,\mathrm{d}x=4+\ln8}