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(x+3)/(x2+2x+3)^2的不定积分

x^2+2x+3 = (x+1)^2 +2 let x+1 =√2tanu dx = √2(secu)^2 .du ----------- ∫ (x+3)/(x^2+2x+3)^2 dx =(1/2) ∫ (2x+2)/(x^2+2x+3)^2 dx + 2∫ dx/(x^2+2x+3)^2 =-(1/2) [1/(x^2+2x+3)] +2∫ dx/(x^2+2x+3)^2 =-(1/2) [1/(x^2+2x+3)] +2∫ √2(secu)^2...

主要步骤如上图。

设(贰x²-三x-三)/[(x-依)(x²-贰x+5)]=[a/(x-依)]+[(bx+c)/(x²-贰x+5)] 则贰x²-三x-三=a(x²-贰x+5)+(bx+c)(x-依) 整理 贰x²-三x-三=(a+b)x²-(贰a+b-c)x+(5a-c) ∴a+b=贰 贰a+b-c=三 5a-c=-三 解 a=-依 b=三 c=...

∫x³/(x²+2x-3)dx=∫(x³+2x-3x-2x+3)/(x²+2x-3)dx =∫x+3/(x²+2x-3)dx =∫xdx+3∫1/(x²+2x-3)dx =x²/2+3∫1/[(x-1)(x+1)]dx =x²/2+3/4∫1/(x-1)-1/(x+3)]dx = x²/2+3/4ln|x-1|-3/4ln|x+3|+C

给了你方法,请自己验算一下,我对着电脑边做边录,不能保证结果无误。

这道题可以直接用三角换元,最后是一个关于θ有关的式子,还原即可

手机照的可能不是太清楚

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

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