cho minh dap an tich phan nay

D

dien0709

$I=\int_{1}^{2} \dfrac{1+x^2}{x^3}lnx dx$

$f(x)dx=\dfrac{1}{x}lnxdx+\dfrac{1}{x^3}lnxdx=udu+I_1$

$u=lnx$=>$du=\dfrac{1}{x}dx$ , $dv=x^{-3}dx$=>$v=-\dfrac{1}{2x^2}$

=>$I_1=-\dfrac{1}{2x^2}lnx-\dfrac{1}{4x^2}$=>I=...
 
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