题目

Problem

Use algebraic integration and the substitution $u=\sqrt{x}$ to find the exact value of

\int_1^4 \frac{10}{5x+2x\sqrt{x}}\,\mathrm{d}x

Write your answer in the form $4\ln\left(\dfrac{a}{b}\right)$ , where $a$ and $b$ are integers to be found.

(Solutions relying entirely on calculator technology are not acceptable.)

(8)

题目中文翻译

使用代数积分，并令 $u=\sqrt{x}$ ，求下式的精确值：

\int_1^4 \frac{10}{5x+2x\sqrt{x}}\,\mathrm{d}x

将答案写成 $4\ln\left(\dfrac{a}{b}\right)$ 的形式，其中 $a,b$ 为待求整数。

（完全依赖计算器技术的解法不接受。）

(8)

解答

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按题意令 $u=\sqrt{x}$ ，把 $x$ 、 $\mathrm{d}x$ 和积分上下限全部改写为 $u$ 。所得有理式可拆成两个简单分式，积分后代入新上下限，再用对数运算法则整理成指定形式。

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Let

u=\sqrt{x}.

Then $x=u^2$ and $\mathrm{d}x=2u\,\mathrm{d}u$ . The limits become

\begin{align*} &\,x=1\Rightarrow u=1,\\ &\,x=4\Rightarrow u=2. \end{align*}

Therefore,

I=\int_1^4\frac{10}{5x+2x\sqrt{x}}\,\mathrm{d}x.

Then

\begin{align*} I =&\,\int_1^2 \frac{10}{5u^2+2u^3}(2u)\,\mathrm{d}u\\ =&\,\int_1^2\frac{20}{u(5+2u)}\,\mathrm{d}u. \end{align*}

Using partial fractions,

write

\begin{align*} &\,\frac{20}{u(5+2u)}\\ =&\,\frac{A}{u}+\frac{B}{5+2u}. \end{align*}

Multiplying by $u(5+2u)$ gives

20=A(5+2u)+Bu.

Setting $u=0$ gives $20=5A$ , so $A=4$ . Comparing the coefficients of $u$ then gives

\begin{align*} 2A+B=&\,0,\\ B=&\,-2(4)=-8. \end{align*}

Thus,

\begin{align*} \frac{20}{u(5+2u)} =&\,\frac{4}{u}\\ -&\,\frac{8}{5+2u}. \end{align*}

Hence,

\begin{align*} I =&\,\left[4\ln u-4\ln(5+2u)\right]_1^2\\ =&\,4\ln2-4\ln9+4\ln7\\ =&\,\boxed{4\ln\left(\frac{14}{9}\right)}. \end{align*}

Thus $a=14$ and $b=9$ .

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仍按题意令 $u=\sqrt{x}$ ，使用解法一求出的部分分式，但先求不定积分，再把 $u$ 换回 $\sqrt{x}$ 。最后代入原来的上下限 $x=1,4$ ，同样能得到所需的精确值。

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With $u=\sqrt{x}$ , we have $x=u^2$ and $\mathrm{d}x=2u\,\mathrm{d}u$ . Using the partial fractions found in the first method,

\begin{align*} &\,\int\frac{10}{5x+2x\sqrt{x}}\,\mathrm{d}x\\ =&\,\int\frac4u\,\mathrm{d}u\\ &\,\hspace{2pt}-\int\frac8{5+2u}\,\mathrm{d}u\\ =&\,4\ln u\\ &\,\hspace{2pt}-4\ln(5+2u)+C\\ =&\,4\ln\sqrt{x}\\ &\,\hspace{2pt}-4\ln(5+2\sqrt{x})+C. \end{align*}

We now apply the original limits $x=1$ and $x=4$ :

\begin{align*} I =&\,\left[\begin{aligned} &\,4\ln\sqrt{x}\\ &\,-4\ln(5+2\sqrt{x}) \end{aligned}\right]_1^4\\ =&\,\big(4\ln2-4\ln9\big)\\ &\,\hspace{2pt}-\big(4\ln1-4\ln7\big)\\ =&\,4\ln2-4\ln9+4\ln7\\ =&\,4\ln\bigg(\frac{2\times7}{9}\bigg)\\ =&\,\boxed{4\ln\bigg(\frac{14}{9}\bigg)}. \end{align*}

Hence $a=14$ and $b=9$ .