第1个回答 2017-12-12
如图
第二题呢？
第2个回答 2017-12-12
(1)
∫ dx/[x^2.√(x^2+1)]
let
x=tanu
dx =(secu)^2 du
∫ dx/[x^2.√(x^2+1)]
=∫ (secu)^2 du/[(tanu)^2.secu]
=∫ [secu /(tanu)^2] du
=∫ [cosu /(sinu)^2] du
=∫ dsinu /(sinu)^2
=-1/sinx +C
=-√(x^2+1)/x + C
(2)
∫ x^2/(1+x^2)^2 dx
=∫ [1/(1+x^2) - 1/(1+x^2)^2 ]dx
=arctanx - ∫ dx/(1+x^2)^2
=arctanx - (1/2)[ arctanx + x/(1+x^2) ] +C
---------
let
x=tanu
dx =(secu)^2 du
∫ dx/(1+x^2)^2
=∫ (cosu)^2 du
=(1/2)∫ (1+cos2u) du
=(1/2)[ u +(1/2)sin2u] +C'
=(1/2)[ arctanx + x/(1+x^2) ] +C'本回答被网友采纳
∫ dx/[x^2.√(x^2+1)]
let
x=tanu
dx =(secu)^2 du
∫ dx/[x^2.√(x^2+1)]
=∫ (secu)^2 du/[(tanu)^2.secu]
=∫ [secu /(tanu)^2] du
=∫ [cosu /(sinu)^2] du
=∫ dsinu /(sinu)^2
=-1/sinx +C
=-√(x^2+1)/x + C
(2)
∫ x^2/(1+x^2)^2 dx
=∫ [1/(1+x^2) - 1/(1+x^2)^2 ]dx
=arctanx - ∫ dx/(1+x^2)^2
=arctanx - (1/2)[ arctanx + x/(1+x^2) ] +C
---------
let
x=tanu
dx =(secu)^2 du
∫ dx/(1+x^2)^2
=∫ (cosu)^2 du
=(1/2)∫ (1+cos2u) du
=(1/2)[ u +(1/2)sin2u] +C'
=(1/2)[ arctanx + x/(1+x^2) ] +C'本回答被网友采纳
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