∫1/(a+bx+cx^2)dx = 1/(√b^2-4ac)ln∣(2cx+b-√b^2-4ac)/2cx+b+√b^2-4ac)∣+C (b^2>4ac)
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∫1/(a+bx+cx^2)dx = 1/(√b^2-4ac)ln∣(2cx+b-√b^2-4ac)/2cx+b+√b^2-4ac)∣+C (b^2>4ac)
∫1/(a+bx+cx^2)dx = 1/(√b^2-4ac)ln∣(2cx+b-√b^2-4ac)/2cx+b+√b^2-4ac)∣+C (b^2>4ac)
∫1/(a+bx+cx^2)dx = 1/(√b^2-4ac)ln∣(2cx+b-√b^2-4ac)/2cx+b+√b^2-4ac)∣+C (b^2>4ac)
当b^2>4ac时,a+bx+cx^2=0有两个根x1,x2,那么a+bx+cx^2=c(x-x1)(x-x2).
令δ=b^2-4ac,两个根是(-b+√δ)/(2c)和(-b-√δ)/(2c).
a+bx+cx^2=c[x+(b-√δ)/(2c)][x+(b+√δ)/(2c)]
原式=1/c∫1/{[x+(b-√δ)/(2c)][x+(b+√δ)/(2c)]}dx=1/c∫c/√δ{1/[(x+(b-√δ)/(2c)]-1/[x+(b+√δ)/(2c)]}dx【注:裂项.1/[(x-x1)(x-x2)]=[1/(x-x1)-1/(x-x2)]*[1/(x1-x2)]】=1/√δ{∫1/[(x+(b-√δ)/(2c)]dx-∫1/[x+(b+√δ)/(2c)]dx}
前一个积分=ln|x+(b-√δ)/(2c)|,后一个积分=ln|x+(b+√δ)/(2c)|
原式=1/√δ【ln|x+(b-√δ)/(2c)|-ln|x+(b+√δ)/(2c)|】=1/√δln|[x+(b-√δ)/(2c)]/[x+(b+√δ)/(2c)]|
化简代入δ即可得公式.(PS:为了省事我没加常数C.)