- 求不定积分(1/跟号(1+x^2))dx答案为ln(x+跟号(1
- 答案为ln(x+跟号(1+x^2)) 求解题过程
- x=tanA
==> cosA =1/根号(1+x^2)
==> 1/根号(1+x^2) =1/根号[1+(tanA)^2] =cosA
积分[1/根号(1+x^2)]dx =积分[cosA*d(tanA)]
= 积分{[d(sinA)]/[1 -(sinA)^2]}
= (1/2)ln[(1+sinA)/(1-sinA)] +C
= (1/2)ln[(1+sinA)^2/(cosA)^2] +C
= ln(tanA +1/cosA) +C
= ln(x+根号(1+x^2)] +C