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∫1/[x+√(1-x²)]dx

高老师2年前 (2024-03-27)高等数学（工专）(00022)7

∫1/[x+√(1-x²)]dx

令x=sint，则dx=costdt ∫[1/(x+√1-x²)]dx=∫[cost/(sint+cost)]/dt =1/2∫[(sint+cost+cost-sint)/(sint+cost)]dt =1/2∫dt+1/2∫d(sint+cost)/(sint+cost) =1/2t+1/2ln|sint+cost|+C =1/2arcsinx+1/2ln|x+√(1+x²)|+C

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