题目

Problem

(a) Differentiate $x\,\operatorname{arcosh}(5x)$ with respect to $x$

(2)

(b) Hence, or otherwise, show that

\int_{1/4}^{3/5}\operatorname{arcosh}(5x)\,\mathrm{d}x =\frac{3}{20}-\frac{2\sqrt2}{5}+\ln(p+q\sqrt2)^{\,k}-\frac14\ln r

where $p$ , $q$ , $r$ and $k$ are rational numbers to be determined.

(8)

题目中文翻译

(a) 对 $x\,\operatorname{arcosh}(5x)$ 关于 $x$ 求导

(2)

(b) 因此，或者用其他方法，证明

\int_{1/4}^{3/5}\operatorname{arcosh}(5x)\,\mathrm{d}x =\frac{3}{20}-\frac{2\sqrt2}{5}+\ln(p+q\sqrt2)^{\,k}-\frac14\ln r

其中 $p$ 、 $q$ 、 $r$ 和 $k$ 为待定有理数。

(8)

解答

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使用乘积法则，并结合 $\frac{\mathrm{d}}{\mathrm{d}x}\operatorname{arcosh}(u)=\frac{u'}{\sqrt{u^2-1}}$ 。这里内层函数是 $u=5x$ ，求导时不要漏掉因子 5。

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Using the product rule,

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \bigl[x\operatorname{arcosh}(5x)\bigr] =&\,\operatorname{arcosh}(5x) \\ &\,+x\cdot\frac{5}{\sqrt{(5x)^2-1}} \\ =&\,\boxed{ \operatorname{arcosh}(5x) +\frac{5x}{\sqrt{25x^2-1}} }. \end{align*}

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将 (a) 的导数公式移项，便可把被积函数表示成一个乘积的导数减去根式积分。求出原函数后代入上下限，再把两个 $\operatorname{arcosh}$ 值化为对数，最后严格整理成题目指定的形式并读出 $p,q,r,k$ 。

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From part (a),

\operatorname{arcosh}(5x) =\frac{\mathrm{d}}{\mathrm{d}x} \bigl[x\operatorname{arcosh}(5x)\bigr] -\frac{5x}{\sqrt{25x^2-1}}.

For the remaining integral, put $w=25x^2-1$ , so $\mathrm{d}w=50x\,\mathrm{d}x$ . Then

\begin{align*} \int\frac{5x}{\sqrt{25x^2-1}}\,\mathrm{d}x =&\,\frac1{10}\int w^{-1/2}\,\mathrm{d}w \\ =&\,\frac15\sqrt{w}+C \\ =&\,\frac15\sqrt{25x^2-1}+C. \end{align*}

Therefore,

\begin{align*} \int\operatorname{arcosh}(5x)\,\mathrm{d}x =&\,x\operatorname{arcosh}(5x) \\ &\,-\int\frac{5x}{\sqrt{25x^2-1}}\,\mathrm{d}x \\ =&\,x\operatorname{arcosh}(5x) -\frac15\sqrt{25x^2-1}+C. \end{align*}

Hence

\begin{align*} \int_{1/4}^{3/5} \operatorname{arcosh}(5x)\,\mathrm{d}x =&\,\biggl[ x\operatorname{arcosh}(5x) -\frac15\sqrt{25x^2-1} \biggr]_{1/4}^{3/5} \\ =&\,\frac35\operatorname{arcosh}(3) -\frac{2\sqrt2}{5} \\ &\,-\frac14\operatorname{arcosh}\biggl(\frac54\biggr) +\frac{3}{20}. \end{align*}

For $t\ge1$ ,

\operatorname{arcosh}(t) =\ln\bigl(t+\sqrt{t^2-1}\bigr).

Thus

\begin{align*} \operatorname{arcosh}(3) =&\,\ln\bigl(3+\sqrt8\bigr) \\ =&\,\ln\bigl(3+2\sqrt2\bigr), \end{align*}

and

\begin{align*} \operatorname{arcosh}\biggl(\frac54\biggr) =&\,\ln\biggl( \frac54+\sqrt{\frac{25}{16}-1} \biggr) \\ =&\,\ln\biggl(\frac54+\frac34\biggr) \\ =&\,\ln2. \end{align*}

Consequently,

\begin{align*} \int_{1/4}^{3/5} \operatorname{arcosh}(5x)\,\mathrm{d}x =&\,\frac{3}{20}-\frac{2\sqrt2}{5} \\ &\,+\frac35\ln\bigl(3+2\sqrt2\bigr) -\frac14\ln2 \\ =&\,\frac{3}{20}-\frac{2\sqrt2}{5} \\ &\,+\ln\bigl(3+2\sqrt2\bigr)^{3/5} -\frac14\ln2. \end{align*}

Comparing this with the required form gives

\boxed{ p=3, \qquad q=2, \qquad r=2, \qquad k=\frac35 }.