x=tant, 则 dx=d(tant)=(sect)^2dt，与原式子(1+（tanx)^2)约掉了。
注意：1+（tanx)^2=(sect）^2
∫(0,π/4) ln2dx -∫(0,π/4) ln(1+tanx)dx
因为该式子就是∫(0,π/4) ln(1+tanx)dx转化而来，所以，
∫(0,π/4) ln2dx -∫(0,π/4) ln(1+tanx)dx=∫(0,π/4) ln(1+tanx)dx
那么：∫(0,π/4) ln2dx =2∫(0,π/4) ln(1+tanx)dx
则：∫(0,π/4) ln(1+tanx)dx=1/2*∫(0,π/4) ln2dx =1/2*ln2*（π/4-0）=π/8*ln2
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第1个回答 2019-07-19
x=tant
dx= (sect)^2 dt
x=0, t=0
x=1, t=π/4
∫(0->1) ln(1+x)/(1+x^2) dx
=∫(0->π/4) [ ln(1+tant)/(1+(tant)^2) ] . [ (sect)^2 dt ]
=∫(0->π/4) ln(1+tant) dt
let
u=π/4-t
du =-dt
t=0, u=π/4
t=π/4, u=0
∫(0->π/4) ln(1+tant) dt
=∫(π/4->0) ln[1+tan(π/4-u) ] (-du)
=∫(0->π/4) ln[1+tan(π/4-u) ] du
=∫(0->π/4) ln[1+ (1-tanu)/(1+tanu) ] du
=∫(0->π/4) ln[2/(1+tanu) ] du
=∫(0->π/4) [ln2 -ln(1+tanu) ] du
=∫(0->π/4) [ln2 -ln(1+tant) ] dt
2∫(0->π/4) ln(1+tant) dt =ln2.∫(0->π/4) dt
∫(0->π/4) ln(1+tant) dt
=(π/8) (ln2)
dx= (sect)^2 dt
x=0, t=0
x=1, t=π/4
∫(0->1) ln(1+x)/(1+x^2) dx
=∫(0->π/4) [ ln(1+tant)/(1+(tant)^2) ] . [ (sect)^2 dt ]
=∫(0->π/4) ln(1+tant) dt
let
u=π/4-t
du =-dt
t=0, u=π/4
t=π/4, u=0
∫(0->π/4) ln(1+tant) dt
=∫(π/4->0) ln[1+tan(π/4-u) ] (-du)
=∫(0->π/4) ln[1+tan(π/4-u) ] du
=∫(0->π/4) ln[1+ (1-tanu)/(1+tanu) ] du
=∫(0->π/4) ln[2/(1+tanu) ] du
=∫(0->π/4) [ln2 -ln(1+tanu) ] du
=∫(0->π/4) [ln2 -ln(1+tant) ] dt
2∫(0->π/4) ln(1+tant) dt =ln2.∫(0->π/4) dt
∫(0->π/4) ln(1+tant) dt
=(π/8) (ln2)
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