CalcGospel 國際數學圖譜
返回

P1 2025 Oct A Q5

A Level / Edexcel / P1

IAL 2025 Oct A Paper · Question 5

题目

Problem

Find

∫4x−32x2 dx,x>0\begin{align*} \int \frac{4\sqrt{x}-3}{2x^2}\,\mathrm{d}x, \qquad x>0 \end{align*}

writing the answer in its simplest form.

(5)
题目中文翻译

求

∫4x−32x2 dx,x>0,\begin{align*} \int \frac{4\sqrt{x}-3}{2x^2}\,\mathrm{d}x, \qquad x>0, \end{align*}

并将答案化为最简形式。（5 分）

解答

解法一

思路

展开

先把分式拆成两项，并把根式写成指数形式。这样就可以直接使用幂函数积分公式。

答题过程

展开

Rewrite the integrand:

4x−32x2= 4x1/22x2−32x2= 2x−3/2−32x−2.\begin{align*} \frac{4\sqrt{x}-3}{2x^2} =&\,\frac{4x^{1/2}}{2x^2}-\frac{3}{2x^2}\\ =&\,2x^{-3/2}-\frac32x^{-2}. \end{align*}

Therefore

∫4x−32x2 dx= ∫(2x−3/2−32x−2) dx= 2(x−1/2−1/2)−32(x−1−1)+c= −4x−1/2+32x−1+c.\begin{align*} \int \frac{4\sqrt{x}-3}{2x^2}\,\mathrm{d}x =&\,\int\left(2x^{-3/2}-\frac32x^{-2}\right)\,\mathrm{d}x\\ =&\,2\left(\frac{x^{-1/2}}{-1/2}\right) -\frac32\left(\frac{x^{-1}}{-1}\right)+c\\ =&\,-4x^{-1/2}+\frac32x^{-1}+c. \end{align*}

So the answer is

−4x+32x+c.\begin{align*} -\frac{4}{\sqrt{x}}+\frac{3}{2x}+c. \end{align*}