求不定积分:∫xdx/ (x^2-2x-3)

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求不定积分:∫xdx/ (x^2-2x-3)
求不定积分:∫xdx/ (x^2-2x-3)

求不定积分:∫xdx/ (x^2-2x-3)
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

把分母分解公因式,(x-3)X(x+1),然后用公式把这个积分转化为两个积分相加的模式就好求了,公式我忘记了,两年没学高数了

x/(x^2-2x-3)
=x/(x-3)(x+1)
=A/(x-3)-B/(x+1)
Ax+A-Bx+3B=x
A-B=1 A+3B=0
A=-5/4 B=-1/4
A/(x-3)-B/(x+1)=1/4(x+1)-5/4(x-3)
所以原式=∫d(x+1)/4(x+1)-∫5d(x-3)/4(x-3)
=1/4*ln|x+1|-5/4*ln|x-3|+C