tích phân khó

T

trantien.hocmai

$I=\int_1^e \frac{1+x^2.lnx}{x+x^2.lnx}dx=\int_1^e \frac{x+x^2lnx}{x+x^2lnx}dx-\int_1^e \frac{x-1}{x+x^2lnx}dx$
$=\int_1^e dx-\int_1^e \frac{\frac{1}{x}-\frac{1}{x^2}}{lnx+\frac{1}{x}}dx$
$=e-1-J$
$J=\int_1^e \frac{\frac{1}{x}-\frac{1}{x^2}}{lnx+\frac{1}{x}}dx$
đặt $u=lnx+\frac{1}{x} \rightarrow du=(\frac{1}{x}-\frac{1}{x^2})dx$
thế là xong nhá
 
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