∫(x+1)\x∧2-x-12dx 学霸,我想了很久就是想不出这道题怎么做,你能帮我看一下吗
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∫(x+1)/(x²-x-12)dx
=(1/2)∫[1/(x²-x-12)]d(x²-x-12)+(3/2)∫[1/(x²-x-12)]dx
=(1/2)ln|x²-x-12|+(3/2)∫[1/(x-4)(x+3)]dx
=(1/2)ln|x²-x-12|+(3/2)·(1/7)∫[1/(x-4)-1/(x+3)]dx
=(1/2)ln|x²-x-12|+(3/14)ln|(x-4)/(x+3)|+C
=(1/2)∫[1/(x²-x-12)]d(x²-x-12)+(3/2)∫[1/(x²-x-12)]dx
=(1/2)ln|x²-x-12|+(3/2)∫[1/(x-4)(x+3)]dx
=(1/2)ln|x²-x-12|+(3/2)·(1/7)∫[1/(x-4)-1/(x+3)]dx
=(1/2)ln|x²-x-12|+(3/14)ln|(x-4)/(x+3)|+C
追问
这个……第一步开始就没看懂啊
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