1除以根号下1减x加x的平方 求其不定积分,
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∫1/(1-x+x^2)dx =∫1/[3/4+(x-1/2)^2]dx =3/4∫1/【1+{(x-1/2)/[(3^(1/2))/2]}^2】dx =[(3^(1/2))/2]∫1/【1+{(x-1/2)/[(3^(1/2))/2]}^2】d{(x-1/2)/[(3^(1/2))/2]} 设{(x-1/2)/[(3^(1/2))/2]}=tant,则.=[(3^(1/2))/2]∫1/[1-(tant)^2]dtant =[(3^(1/2))/2]∫[1/(sect)^2]*(sect)dt =[(3^(1/2))/2]∫dt =[(3^(1/2))/2]t+C 由{(x-1/2)/[(3^(1/2))/2]}=tant得.t=arctan【{(x-1/2)/[(3^(1/2))/2]}】带入得.=[(3^(1/2))/2]*arctan【{(x-1/2)/[(3^(1/2))/2]}】+C 如果有不妥请回复.

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