求不定积分：∫xdx/ (x^2-2x-3)_数学解答

求不定积分：∫xdx/ (x^2-2x-3)

人气：381 ℃　时间：2020-06-20 21:04:55

解答

x/(x²-2x-3)
=(x-3+3)/(x-3)(x+1)
=1/(x+1)+3/(x-3)(x+1)
=1/(x+1)+3/4*[1/(x-3)-1/(x+1)]
=3/4*1/(x-3)+1/4*1/(x+1)
所以原式=3/4*ln|x-3|+1/4*ln|x+1|+C