∫2x+1⼀x^2+2X-3 dx. 不定积分的详细过程和答案,拜托大神.

2024-11-08 18:43:29
推荐回答(1个)
回答(1):

分母因式分解为:(x+3)(x-1)
令:(2x+1)/[(x+3)(x-1)]=A/(x+3)+B/(x-1)
右边通分合并,与左边比较系数后得:A=5/4,B=3/4
则:∫ (2x+1)/(x²+2x-3) dx
=(5/4)∫ 1/(x+3) dx + (3/4)∫ 1/(x-1) dx
=(5/4)ln|x+3| + (3/4)ln|x-1| + C
希望可以帮到你,如果解决了问题,请点下面的"选为满意回答"按钮.