题目

Problem

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

Solutions relying entirely on calculator technology are not acceptable.

Given that $k$ is a positive constant,

(a) find

\int \frac{9x}{3x^2+k}\,dx

(2)

Given also that

\int_2^5 \frac{9x}{3x^2+k}\,dx=\ln 8

(b) find the value of $k$

(4)

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。

已知 $k$ 是正常数。

(a) 求

\int \frac{9x}{3x^2+k}\,dx

又已知

\int_2^5 \frac{9x}{3x^2+k}\,dx=\ln 8

(b) 求 $k$ 的值。

解答

We need to find

\int \frac{9x}{3x^2+k}\,\mathrm{d}x

Let

u=3x^2+k

Then

\frac{\mathrm{d}u}{\mathrm{d}x}=6x

9x\,\mathrm{d}x=\frac32\,\mathrm{d}u

Therefore

\int \frac{9x}{3x^2+k}\,\mathrm{d}x =\frac32\int \frac{1}{u}\,\mathrm{d}u

Hence

\boxed{\int \frac{9x}{3x^2+k}\,\mathrm{d}x =\frac32\ln(3x^2+k)+C}

Using part (a),

\int_2^5 \frac{9x}{3x^2+k}\,\mathrm{d}x =\left[\frac32\ln(3x^2+k)\right]_2^5

\frac32\ln(75+k)-\frac32\ln(12+k)=\ln8

Factor out $\dfrac32$ :

\frac32\left[\ln(75+k)-\ln(12+k)\right]=\ln8

Using the logarithm law,

\frac32\ln\left(\frac{75+k}{12+k}\right)=\ln8

Multiply by $\dfrac23$ :

\ln\left(\frac{75+k}{12+k}\right)=\frac23\ln8

Since

\frac23\ln8=\ln(8^{2/3})=\ln4

we have

\ln\left(\frac{75+k}{12+k}\right)=\ln4

Therefore

\frac{75+k}{12+k}=4

75+k=48+4k

Thus

27=3k

and hence

\boxed{k=9}