Question

1、求不定积分1/x^2*sin(2/x+3)dx 2、求由曲面z=2x^2+2y^2及z=6-x^2-y^2所围成的立体体积

3、设f(x)二阶可导,求y=f(根号(x^2+1))对x的一阶导数和二阶导数

Answer 1

1.    ∫1/x^2*sin(2/x+3)dx

   =-∫sin(2/x +3)d(1/x)

   =-1/2 ∫sin(2/x +3)d(2/x+3)

   =1/2 cos(2/x +3)+c

2.  z=6-z/2,

   2z=12-z

   3z=12

   z=4

   x²＋y²=2

   体积=∫∫ (6-x²-y²-2x²-2y²)dxdy

   =∫∫(6-3x²-3y²)dxdy

   =∫∫(6-3p²)pdpdθ

   =∫（0,2π）dθ∫(0,根号2)（6p-3p³）dp

   =2π （3p²-3/4p^4）|(0，根号2)

   =2π （3×2-3/4 ×2²）

   =2π×3

   =6π

3. y'=f'(√(x²＋1))· 2x/[2√(x²＋1)]

   =x/√(x²＋1) ·f'(√(x²＋1))

   y''=(√(x²＋1)-x·x/√(x²＋1))/√(x²＋1)² ·f'(√(x²＋1))+x/√(x²＋1) f''(√(x²＋1))·x/√(x²＋1)

   =1/(x²+1)^(3/2) f'(√(x²＋1)) +x²/(x²＋1) f''(√(x²＋1))