15 ∫xcos2xdx=1/2 ∫xcos2x d(2x)
=1/2 ∫x d(sin2x)
=1/2 [xsin2x -∫sin2xdx] 分部积分法
=1/2 [xsin2x -1/2∫sin2xd(2x)]
=1/2 xsin2x + 1/4 cos2x + C
16 令√1-x =t,x=1-t^2,dx=-2t dt
∫1/(√1-x +1) dx=∫1/(t+1) * (-2t)dt
=-2 ∫[1-1/(t+1)]dt
=-2t + 2ln(t+1) + C
=-2√1-x +2ln(√1-x + 1)+C
代入积分上下限,结果= 2 - 2ln2
13 原式=∫1/[1+(4x)^2 ] dx
=1/4 ∫1/[1+(4x)^2 ] d(4x)
=1/4 arctan4x + C
8 这一题我还没学过,仅供参考.
dy/dx - x/e^y =0
e^y dy = x dx
两边积分得:e^y = x^2/2 + c
y=ln(x^2/2 + c)
只写到这一步,然后你想改写成什么形式随便.
