计算要过程∫（2x+1)/(x^2+4x+3)dx

网友回答

∫（2x+1)/(x^2+4x+3)dx
=∫（2x+1)/[(x+3)(x+1)]dx
=(5/2)∫1/(x+3)dx-(1/2)∫1/(x+1)dx
=(5/2)ln|x+3|-(1/2)ln|x+1|+C
======以下答案可供参考======
供参考答案1:
∫(2x+1)/(x^2+4x+3)dx
设（2x+1)/(x^2+4x+3)=A/(x+1)+B/(x+3)
解得：A=-1/2 B=5/2
所以：∫（2x+1)/(x^2+4x+3)dx
=∫((-1/2)/(x+1)+(5/2)/(x+3))dx
=(-1/2)ln|x+1|+(5/2)ln|x+3|+C
供参考答案2:
∫(2x+1)/(x²+4x+3)dx
=∫(x+x+1)/(x+1)(x+3)dx
=∫[x/(x+1)(x+3)+1/(x+3)]dx
=∫[x/(x²+4x+3)]dx+ln|x+3|+c
=(1/2)ln|x²+4x+3|-(3/2)∫1/(x+1)(x+3)dx+ln|x+3|+c
=(1/2)ln|x²+4x+3|-(3/4)∫[1/(x+1)-1/(x+3)]dx+ln|x+3|+c
=(1/2)ln|x²+4x+3|-(3/4)ln|(x+1)/(x+3)|+ln|x+3|+c