不定积分的典型例题

x + sin x 例 51. ∫ dx = ∫ 1 + cos x

x x x + 2 sin cos 2 2 = 2 x 2 cos 2 x (分项分部积分) = x tan + c. 2

*例 52.求 ∫

f ′(ln x) f ′(ln x) dx = ∫ d (ln x) x f (ln x) f (ln x)

=∫

d ( f (ln x)) = 2 f (ln x) + c. f (ln x)

*例 53.求

∫ max(

x 3 , x 2 ,1 ) dx .

x3 , x ≥ 1 解 令 f ( x) = max( x 3 , x 2 ,1) = x 2 , x ≤ 1 1, x < 1

1 4 4 x + c1 , x ≥ 1 1 max( x 3 , x 2 ,1)dx = x 3 + c2 , x ≤ 1. ∫ 3 x + c3 , x < 1 利用原函数的连续性,有1 lim ( x 4 + c1 ) = lim ( x + c3 ); + x →1 4 x →1 1 lim+ ( x + c3 ) = lim ( x 3 + c2 ), x → 1 x → 1 3 3 2 + c, c2 = c, 4 3 1 4 3 4 x + 4 + c , x ≥1 1 2 故 ∫ max( x 3 , x 2 ,1)dx = x 3 + c , x ≤ 1. 3 3 x + c, x <1

从而解出 c3 = c, c1 =

315