题目

Problem

Given that

\begin{align*} (x-5)(2x+1)(x+3)\equiv ax^3+bx^2-32x-15 \end{align*}

where $a$ and $b$ are constants,

(a) find the value of $a$ and the value of $b$ .

(2)

(b) Hence find

\begin{align*} \int \frac{(x-5)(2x+1)(x+3)}{5\sqrt{x}}\,dx \end{align*}

writing each term in simplest form.

(5)

解答

展开

直接展开左边即可。也可以只找 $x^3$ 和 $x^2$ 的系数，但完整展开能顺便为下一小题服务。

展开

First multiply two brackets:

\begin{align*} (x-5)(x+3) =&\,x^2-2x-15. \end{align*}

Then

\begin{align*} (x-5)(2x+1)(x+3) =&\,(2x+1)(x^2-2x-15)\\ =&\,2x^3-4x^2-30x\\ &\quad +x^2-2x-15\\ =&\,2x^3-3x^2-32x-15. \end{align*}

Therefore

\begin{align*} a=2, \qquad b=-3. \end{align*}

展开

用上一小题的展开式，把分子换成多项式。分母是 $5\sqrt{x}=5x^{1/2}$ ，所以每一项分别除以 $5x^{1/2}$ ，再逐项积分。

展开

Using part (a),

\begin{align*} \frac{(x-5)(2x+1)(x+3)}{5\sqrt{x}} =&\,\frac{2x^3-3x^2-32x-15}{5x^{\frac12}}\\ =&\,\frac25x^{\frac52}-\frac35x^{\frac32}\\ &\quad -\frac{32}{5}x^{\frac12}-3x^{-\frac12}. \end{align*}

Therefore

\begin{align*} \int \frac{(x-5)(2x+1)(x+3)}{5\sqrt{x}}\,dx =&\,\int \left( \frac25x^{\frac52}-\frac35x^{\frac32}\right.\\ &\quad \left.-\frac{32}{5}x^{\frac12} -3x^{-\frac12} \right)\,dx\\ =&\,\frac4{35}x^{\frac72}-\frac6{25}x^{\frac52}\\ &\quad -\frac{64}{15}x^{\frac32} -6x^{\frac12}+c. \end{align*}