题目

Problem

(i) Find, using algebraic integration, the exact value of

\int_3^4 \frac{42}{(3x-1)^2}\,dx

giving your answer in simplest form.

(4)

(ii)

h(x)=\frac{2x^3-7x^2+8x+1}{(x-1)^2}\qquad x>1

Given $h(x)=Ax+B+\dfrac{C}{(x-1)^2}$ where $A$ , $B$ and $C$ are constants to be found, find

\int h(x)\,dx

(6)

题目中文翻译

(i) 利用代数积分法求

\int_3^4 \frac{42}{(3x-1)^2}\,dx

的精确值，并将答案化为最简形式。

(ii)

h(x)=\frac{2x^3-7x^2+8x+1}{(x-1)^2}\qquad x>1

已知

h(x)=Ax+B+\dfrac{C}{(x-1)^2}

其中 $A$ 、 $B$ 和 $C$ 为待求常数，求

\int h(x)\,dx

解答

We need to find

\int_3^4 \frac{42}{(3x-1)^2}\,\mathrm{d}x

Write the integrand using a negative power:

\frac{42}{(3x-1)^2}=42(3x-1)^{-2}

Since the derivative of $3x-1$ is $3$ ,

\int 42(3x-1)^{-2}\,\mathrm{d}x =42\cdot \frac{(3x-1)^{-1}}{-1}\cdot \frac13

\int 42(3x-1)^{-2}\,\mathrm{d}x =-14(3x-1)^{-1} =-\frac{14}{3x-1}

Therefore

\begin{aligned} \int_3^4 \frac{42}{(3x-1)^2}\,\mathrm{d}x &=\left[-\frac{14}{3x-1}\right]_3^4\\ &=-\frac{14}{11}-\left(-\frac{14}{8}\right)\\ &=-\frac{14}{11}+\frac74\\ &=\frac{-56+77}{44}\\ &=\frac{21}{44} \end{aligned}

Hence

\boxed{\frac{21}{44}}

We are given

h(x)=\frac{2x^3-7x^2+8x+1}{(x-1)^2}

and

h(x)=Ax+B+\frac{C}{(x-1)^2}

Multiply by $(x-1)^2$ :

2x^3-7x^2+8x+1=(Ax+B)(x-1)^2+C

Expand:

(Ax+B)(x^2-2x+1)+C

2x^3-7x^2+8x+1 =Ax^3+(B-2A)x^2+(A-2B)x+(B+C)

Compare coefficients:

A=2

Then

B-2A=-7

B-4=-7

and therefore

B=-3

Now check the coefficient of $x$ :

A-2B=2-2(-3)=8

which agrees with the numerator.

For the constant term:

B+C=1

-3+C=1

and therefore

C=4

Thus

h(x)=2x-3+\frac{4}{(x-1)^2}

Now integrate:

\begin{aligned} \int h(x)\,\mathrm{d}x &=\int \left(2x-3+4(x-1)^{-2}\right)\,\mathrm{d}x\\ &=x^2-3x+4\cdot \frac{(x-1)^{-1}}{-1}\\ &=x^2-3x-\frac{4}{x-1}+c \end{aligned}

Therefore

\boxed{\int h(x)\,\mathrm{d}x=x^2-3x-\frac{4}{x-1}+c}