∫x/(x的平方+2x+3)dx
=∫x/[(x+1)²＋2] dx
=∫[(x+1)-1]/[(x+1)²＋2] dx
=∫[(x+1)/[(x+1)²＋2] dx-∫1/[(x+1)²＋2] dx
=1/2∫1/(x的平方+2x+3) d(x的平方+2x+3) -1/√2 arctan(x+1)/√2+c
=1/2ln(x的平方+2x+3)-√2/2 arctan(x+1)/√2+c1/2∫1/(x的平方+2x+3) d(x的平方+2x+3) -1/√2 arctan(x+1)/√2+c这步怎么来的啊，请详解