求不定积分：1、∫1/[x^2(x^2+1)]dx 2、∫sinx/(1+sinx）dx

1、∫1/[x²(x²+1)]dx =∫[1/x²-1/(x²+1)]dx=∫dx/x²-∫dx/(x²+1)=-1/x-arctanx+C (C是积分常数)2、∫sinx/(1+sinx)dx=∫(1+sinx-1)/(1+sinx)dx=∫[1-1/(1+sinx)]dx=∫dx-∫dx/(1+sinx)=x-∫dx/[sin²(x/2)+cos²(x/2)+2sin(x/2)cos(x/2)]=x-∫dx/[sin(x/2)+cos(x/2)]²=x-∫sec²(x/2)/[tan(x/2)+1]²dx=x-∫d[tan(x/2)]/[tan(x/2)+1]²dx=x-∫d[tan(x/2)+1]/[tan(x/2)+1]²dx=x+1/[tan(x/2)+1]+C (C是积分常数)