Question

请教高数高手：求1/(1+2*tanx)的不定积分

Answer 1

∫ dx/(1+2tanx)
= ∫ cosx/(cosx+2sinx) dx
= (1/5)∫ dx + (2/5)∫ (2cosx-sinx)/(cosx+2sinx) dx
= (1/5)x + (2/5)∫ d(cosx+2sinx)/(cosx+2sinx) dx
= (1/5)x + (2/5)ln|cosx+2sinx| + C

Note：
cosx = A(cosx+2sinx) + B(2cosx-sinx)
cosx = (A+2B)cosx + (2A-B)sinx
2A-B=0 => B=2A
A+2B=1
A+2(2A)=1
A=1/5
B=2/5

Answer 2

由
cosx = A(cosx+2sinx) + B(2cosx-sinx)
cosx = (A+2B)cosx + (2A-B)sinx
2A-B=0 得 B=2A
A+2B=1
A+2(2A)=1
A=1/5 B=2/5
∫ dx/(1+2tanx)
= ∫ cosx/(cosx+2sinx) dx
= (1/5)∫ dx + (2/5)∫ (2cosx-sinx)/(cosx+2sinx) dx
= (1/5)x + (2/5)∫ 1/(cosx+2sinx) d(cosx+2sinx)
= (1/5)x + (2/5)ln|cosx+2sinx| + C