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IAL 2025 May Q3

A Level / Edexcel / P1

IAL 2025 May Paper · Question 3

题目

Problem

Find

(x+3)23xdx,\begin{align*} \int \frac{(x+3)^2}{3\sqrt{x}}\,dx, \end{align*}

writing your answer in simplest form.

(5)

解答

解法一

思路

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先展开分子,再逐项除以 3x3\sqrt{x}。化成幂函数后就可以逐项积分。

答题过程

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Expand the numerator:

(x+3)2=x2+6x+9.\begin{align*} (x+3)^2=x^2+6x+9. \end{align*}

Then

(x+3)23x=x2+6x+93x1/2=13x3/2+2x1/2+3x1/2.\begin{align*} \frac{(x+3)^2}{3\sqrt{x}} =&\,\frac{x^2+6x+9}{3x^{1/2}}\\ =&\,\frac13x^{3/2}+2x^{1/2}+3x^{-1/2}. \end{align*}

So

(x+3)23xdx=(13x3/2+2x1/2+3x1/2)dx=13x5/25/2+2x3/23/2+3x1/21/2+c=215x5/2+43x3/2+6x1/2+c.\begin{align*} \int \frac{(x+3)^2}{3\sqrt{x}}\,dx =&\,\int\left( \frac13x^{3/2}+2x^{1/2}+3x^{-1/2} \right)\,dx\\ =&\,\frac13\cdot\frac{x^{5/2}}{5/2} +2\cdot\frac{x^{3/2}}{3/2} +3\cdot\frac{x^{1/2}}{1/2}+c\\ =&\,\frac{2}{15}x^{5/2} +\frac43x^{3/2} +6x^{1/2}+c. \end{align*}