∫dx/[(1+x)√1-2]
令x=sint，则dx=costdt于是∫dx/(1+x)√1-x2=∫costdt/(1+sint)cos=∫dt/(1+sint)=∫(1-sint)dt/cos2t=∫sec2tdt+∫(1/cos2t)dcost=tant-(1/cost)+C=x/√1-x2-1/√1-x2+C=-√[(1-x)/(1+x)]+C
∫dx/[(1+x)√1-2]
令x=sint，则dx=costdt于是∫dx/(1+x)√1-x2=∫costdt/(1+sint)cos=∫dt/(1+sint)=∫(1-sint)dt/cos2t=∫sec2tdt+∫(1/cos2t)dcost=tant-(1/cost)+C=x/√1-x2-1/√1-x2+C=-√[(1-x)/(1+x)]+C
