HMMT 二月 2007 · 冲刺赛 · 第 33 题

HMMT February 2007 — Guts Round — Problem 33

[ 18 ] Compute ∫ 2 9 x + 4 dx. 5 2 x + 3 x + x 1 (No, your TI-89 doesn’t know how to do this one. Yes, the end is near.) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 10 HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND

解析

[ 18 ] Compute ∫ 2 9 x + 4 dx. 5 2 x + 3 x + x 1 (No, your TI-89 doesn’t know how to do this one. Yes, the end is near.) 80 Answer: ln . We break the given integral into two pieces: 23 ∫ ∫ ∫ 2 2 2 4 4 9 x + 4 x + 3 x + 1 5 x + 6 x + 1 dx = 5 dx − dx. 5 2 5 2 5 2 x + 3 x + x x + 3 x + x x + 3 x + x 1 1 1 11 These two new integrals are easily computed; for, the first integrand reduces to 1 /x and the second is ′ of the form f ( x ) /f ( x ). We obtain [ ] 80 2 5 2 5 ln | x | − ln | x + 3 x + x | = ln 32 − ln 46 + ln 5 = ln 1 23 5 2 4 Motivation. Writing f ( x ) = 9 x + 4 and g ( x ) = x + 3 x + x = x ( x + 3 x + 1), we wish to find the antiderivative of f ( x ) /g ( x ). It makes sense to consider other rational functions with denominator 4 g ( x ) that have an exact antiderivative. Clearly, if the numerator were f ( x ) = x + 3 x + 1 or a 1 constant multiple, then we can integrate the function. Another trivial case is if the numerator were ′ 4 f ( x ) = g ( x ) = 5 x + 6 x + 1 or a constant multiple. Guessing that f ( x ) is a linear combination of 2 f ( x ) and f ( x ), we easily find that f ( x ) = 9 x + 4 = 5 f ( x ) − f ( x ). 1 2 1 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 10 HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND