求三道不定积分∫√(1+sinx)dx,∫1/(x∧2+4x-5)dx,

题目

求三道不定积分∫√(1+sinx)dx,∫1/(x∧2+4x-5)dx,
求三道不定积分∫√(1+sinx)dx,∫1/(x∧2+4x-5)dx,∫(3x+1)/(x∧2+4x-5)dx

答案

1+sinx=1+cos(π/2-x)=2cos²(π/4-x/2)
1/(x²+4x-5)=1/[(x+5)(x-1)]=[1/(x-1)-1/(x+5)]·1/6
(3x+1)/(x²+4x-5)=[(3x+1)/(x-1)-(3x+1)/(x+5)]·1/6=7/3(x+5)+2/3(x-1)
接下来很容易算