灵鹊做题网|（x^2+1）/((x+1)*(x-2)^2)dx.求不定积分?.求大神指教.
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|（x^2+1）/((x+1)*(x-2)^2)dx.求不定积分?.求大神指教.
先用待定系数法把该有理分式分解为最简分式之和,再行积分.设
(x^2+1)/[(x+1)*(x-2)^2] = A/(x+1) + B/(x-2) + C/(x-2)^2,
去分母,得
x^2+1 = A(x-2)^2 + B(x+1)(x-2) + C(x+1),
对比同次项系数,可求得A,B,C（略,留给你）.
这样,
原积分 = A∫[1/(x+1)]dx + B∫[1/(x-2)]dx + C∫[1/(x-2)^2]dx
= A*ln(x+1) + B*ln(x-2) - C/(x-2) + Const.
(x^2+1)/[(x+1)*(x-2)^2] = A/(x+1) + B/(x-2) + C/(x-2)^2,
去分母,得
x^2+1 = A(x-2)^2 + B(x+1)(x-2) + C(x+1),
对比同次项系数,可求得A,B,C（略,留给你）.
这样,
原积分 = A∫[1/(x+1)]dx + B∫[1/(x-2)]dx + C∫[1/(x-2)^2]dx
= A*ln(x+1) + B*ln(x-2) - C/(x-2) + Const.