x/(3x+4)^2dx的不定积分怎么求呀?求推导过程,公式知道.

Question

x/(3x+4)^2 dx的不定积分怎么求呀?求推导过程,公式知道.

Answer 1

令 x/(3x+4)^2 = A/(3x+4)^2 + B/(3x+4)
=[3Ax + (4A+B)]/(3x+4)^2
则
{
3A=1→A=1/3
4A+B=0→ B= -4/3
即 x/(3x+4)^2 = (1/3)/(3x+4)^2 - (4/3)/(3x+4)
原积分化为
=∫ [(1/3)/(3x+4)^2 - (4/3)/(3x+4)] dx
=(1/3)∫ 1/(3x+4)^2 dx - (4/3)∫ 1/(3x+4) dx
=(1/9)∫ 1/(3x+4)^2 d(3x+4) - (4/9)∫ 1/(3x+4) d(3x+4)
= -(1/9) / (3x+4) - (4/9)ln|3x+4| +C