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

題目：

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

解答：

1、∫1/[x²(x²+1)]dx
=∫[1/x²-1/(x²+1)]dx
=∫dx/x²-∫dx/(x²+1)
=-1/x-arctanx+C (C是積分常數)
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是積分常數)