求下列不定积分.（1）∫[1/(x+1)^2 (x^2+1)]dx (2) ∫[1/(2+sinx)]dx (3) ∫[

答：
1.原式
=∫[(x+2)/[2(x+1)^2]-x/[2(x^2+1)] dx
=1/2*∫[(x+1+1)/(x+1)^2-x/(x^2+1) dx
=1/2*∫(1/(x+1)+1/(x+1)^2-x/(x^2+1))dx
=1/2*[ln|x+1|-1/(x+1)-1/2*ln(x^2+1)]+C
=1/2*ln|(x+1)/√(x^2+1)|-1/2(x+1)+C
2.原式
=1/2*∫1/(1+1/2*sinx)dx
令t=tan(x/2),则sinx=2t/(1+t^2),cosx=(1-t^2)/(1+t^2),dx=2dt/(1+t^2)
=∫dt/(1+t+t^2)
=4/3*∫1/{[(2t+1)/√3]^2+1}
=4/3*√3/2*arctan[(2t+1)/√3]+C
=2/√3*arctan[(2tan(x/2)+1)/√3]+C
上面这个结果可以化简.所以答案形式不唯一.
3.原式
=∫(1+sinx-1)/(1+sinx)dx
=∫1-1/(1+sinx)dx
=∫1-1/(1+cos(x-π/2))dx
由cos2t=2(cost)^2-1可得：
=∫1-1/(1+2[cos(x/2-π/4)]^2-1)dx
=∫1-1/2cos(x/2-π/4)^2 dx
=x-tan(x/2-π/4)+C
化简得：
=x+cosx/(1+sinx)+C
注：第二题也可以令sinx=cos(x-π/2),然后按照《吉米多维奇数学分析习题集》(第三版)第2028题解法.