北大版高等数学第二章_微积分的基本概念答案_习题2.5

∫ (ax − be
−x
+源自文库C1 )dx =
解( xf ( x ))′ = x 2 + 1, xf ( x ) = f ( x) = 1 3 C x + 1+ . 4 x
∫x
3
+ 1dx =
1 4 x + x + C, 4
2
x
∫ 2
x
x x 1 1 1 + ÷ ÷ ÷dx 3 9 2 ÷
1 1 1 = − 2 ÷ / ln 3 − ÷ / ln 2 + C. 9 2 3 1 1 1 1 16.∫ 2 dx = ∫ 2 − dx = − − arctan x + C. 2 2 ÷ x (1 + x ) x x 1+ x
3 7.∫ + x
2 3 1− x 4 5 / 4 2 3/ 2 4 7 / 4 4 4 4 dx = (1 + x + x + x ) dx = x + x + x + x +C ∫ 5 3 7 1− 4 x 2 6 1 2 3 10.∫ + 2 + 3 ÷dx = ln | x | − − 2 + C. x x x x x (1 − x) 2 1 − 2 x + x2 11.∫ dx = dx = ∫ ( x − 4 / 3 − 2 x − 1/ 3 + x 2 / 3 ) dx 4/ 3 ∫ 3 x xg x 3 = − 3x − 1/ 3 − 3x 2/ 3 + x5/ 3 + C. 5
1
17.求解微分方程y′′ ( x) = a + be − x (a, b为常数). 解y ′ = ∫ (a + be− x )dx = ax − be − x + C1 , y 1 2 ax + be − x + C1x + C2 . 2 ′ 18.设f ( x)满足方程xf ( x) + f ( x) = x 2 + 1, 求f ( x). =
8.∫ (1 + cos 2 x)sec2 xdx = ∫ (sec 2 x + 1)dx = tan x + x + C. 9.∫
12.∫ (2cosh x − sinh x)dx = 2sinh x − cosh x + C. 3x 2 − 1 ( x + 1) 2 1 3/ 2 1/ 2 − 1/ 2 13.∫ + ÷dx = ∫ 3 − 2 + x + 2 x + x ÷dx 2 x x x 1 2 4 = 3x + + x5/ 2 + x3/ 2 + 2 x + C. x 5 3 1 1 1 14.∫ dx = ∫ + ÷dx = − cot x + tan x + C. 2 2 2 2 sin x cos x sin cos x 15.∫ 2 x + 1 + 3x − 2 dx = 6x
习题 2.5
求下列不定积分 : b b 9C 5/ 3 a 1.∫ − 2 + 3C 3 x 2 ÷dx = 2a x + + x + C. x 5 x x 2 4 1 2.∫ (1 + x ) dx = ∫ (1 + 2 x + x)dx = x + x3/ 2 + x 2 + C. 3 2 3.∫ a sec 2 xdx = a tan x + C. 4.∫ tan 2 xdx = ∫ (sec 2 x − 1)dx = tan x − x + C. 5.∫ cot 2ϕ dϕ = ∫ (csc2ϕ − 1) dϕ = − cot 2 ϕ − 1 + C. 6.∫ x2 + 3 2 dx = ∫ 1 + dx = x + 2arctan x + C. 2 2 ÷ 1+ x 1+ x ÷dx = 6 x + 4arcsin x + C. 1 − x2 4