Question

求解：求下列不定积分(1)∫(x+2)/(x²+3x+4)dx;(2)1/√(1-2x-x²)

Answer 1

∫(x+2)/(x²+3x+4) dx
=(1/2)∫(2x+3)/(x²+3x+4) dx + (1/2)∫dx/(x²+3x+4)
=(1/2)∫d(x²+3x+4)/(x²+3x+4) + (1/2)∫dx/[(x+3/2)²+(√7/2)²]
=(1/2)ln|x²+3x+4| + (1/2)/(√7/2)*∫d[(x+3/2)/(√7/2)] / [1+(x+3/2)²/(√7/2)²]，∫dx/(1+x²)形式
=(1/2)ln|x²+3x+4| + (1/√7)arctan[(x+3/2)/(√7/2)] + C
=(1/2)ln|x²+3x+4| + (1/√7)arctan[(2x+3)/√7] + C

∫dx/√(1-2x-x²)
=∫dx/√[2-(x+1)²]
=∫d[(x+1)/√2]/√[1-(x+1)²/(√2)²]，∫dx/√(1-x²)形式
=arcsin[(x+1)/√2] + C