∫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
∫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
