HMMT 二月 2009 · CALC 赛 · 第 3 题

HMMT February 2009 — CALC Round — Problem 3

[ 4 ] Compute e where A is defined as ∫ 4 / 3 2 2 x + x + 1 dx. 3 2 x + x + x + 1 3 / 4 ′ ′

解析

[ 4 ] Compute e where A is defined as ∫ 4 / 3 2 2 x + x + 1 dx. 3 2 x + x + x + 1 3 / 4 16 Answer: 9 1 x Solution: We can use partial fractions to decompose the integrand to + , and then integrate 2 x +1 x +1 2 the addends separately by substituting u = x + 1 for the former and u = x + 1 for latter, to obtain ∣ √ ∣ 4 / 3 4 / 3 1 16 2 A 2 ∣ ∣ ln( x + 1) + ln( x + 1) = ln(( x + 1) x + 1) = ln . Thus e = 16 / 9. 2 9 3 / 4 3 / 4 Alternate solution: Substituting u = 1 /x , we find ∫ ∫ 3 / 4 4 / 3 2 3 2 u + u + u 1 2 /u + 1 + u A = ( − ) du = du 2 3 2 2 3 1 + u + u + u u 1 + u + u + u 4 / 3 3 / 4 Adding this to the original integral, we find ∫ ∫ 4 / 3 4 / 3 2 2 /u + 2 + 2 u + 2 u 2 2 A = du = du 2 3 1 + u + u + u u 3 / 4 3 / 4 16 16 A Thus A = ln and e = . 9 9 1 ′ ′