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IAL 2021 Jan Q1

A Level / Edexcel / P3

IAL 2021 Jan Paper · Question 1

题目

Problem

Find

∫x2−52x3 dxx>0\int \frac{x^2-5}{2x^3}\,dx \qquad x>0

giving your answer in simplest form.

(3)
题目中文翻译

求

∫x2−52x3 dxx>0\int \frac{x^2-5}{2x^3}\,dx \qquad x>0

并将答案化为最简形式。

解答

First simplify the integrand:

x2−52x3=x22x3−52x3\frac{x^2-5}{2x^3} =\frac{x^2}{2x^3}-\frac{5}{2x^3}

So

x2−52x3=12x−52x−3\frac{x^2-5}{2x^3} =\frac{1}{2x}-\frac{5}{2}x^{-3}

Therefore

∫x2−52x3 dx=∫(12x−52x−3) dx\int \frac{x^2-5}{2x^3}\,\mathrm{d}x =\int \left(\frac{1}{2x}-\frac{5}{2}x^{-3}\right)\,\mathrm{d}x

Integrate term by term:

∫12x dx=12ln⁡x\int \frac{1}{2x}\,\mathrm{d}x=\frac12\ln x

because x>0x>0.

Also,

∫−52x−3 dx=−52⋅x−2−2=54x−2\int -\frac52x^{-3}\,\mathrm{d}x =-\frac52\cdot \frac{x^{-2}}{-2} =\frac54x^{-2}

Hence

∫x2−52x3 dx=12ln⁡x+54x2+c\boxed{\int \frac{x^2-5}{2x^3}\,\mathrm{d}x =\frac12\ln x+\frac{5}{4x^2}+c}