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求不定积分:∫xdx/(x^2-2x-3)

题目详情
求不定积分:∫xdx/ (x^2-2x-3)
▼优质解答
答案和解析
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