+>771

∫{1/(x^2)}*log(x)

f`(x)=1/(x^2)
g(x)=log(x)
とおいて、

=-(1/x)log(x)-∫-(1/x)*(1/x)
={-(log(x))/(x)}-∫-x^(-2)
={-(log(x))/(x)}-(-(1/x))
={-(log(x))/(x)}+(1/x)+c

ですか?
なるほど。

∫{1/(x^2)}*log(x)

f`(x)=1/(x^2)
g(x)=log(x)
とおいて、

=-(1/x)log(x)-∫-(1/x)*(1/x)
={-(log(x))/(x)}+log(x)^2

であってますか?