图示不定积分怎么求
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let
x=tanu
dx=(secu)^2 du
x=1, u=π/4
x=√3, u=π/3
∫(1->√3) dx/[x^2.√(1+x^2)]
=∫(π/4->π/3) (secu)^2 du/[(tanu)^2.secu]
=∫(π/4->π/3) [(secu)/(tanu)^2] du
=∫(π/4->π/3) [ cosu/(sinu)^2] du
=∫(π/4->π/3) dsinu/(sinu)^2
= -[1/sinu]|(π/4->π/3)
=√2 - (2/3)√3
x=tanu
dx=(secu)^2 du
x=1, u=π/4
x=√3, u=π/3
∫(1->√3) dx/[x^2.√(1+x^2)]
=∫(π/4->π/3) (secu)^2 du/[(tanu)^2.secu]
=∫(π/4->π/3) [(secu)/(tanu)^2] du
=∫(π/4->π/3) [ cosu/(sinu)^2] du
=∫(π/4->π/3) dsinu/(sinu)^2
= -[1/sinu]|(π/4->π/3)
=√2 - (2/3)√3
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