Question

1->dương vô cùng 1/x*(9+lnx^2)dx

Answer 1

It looks like you are trying to compute the improper integral,

`I = \displaystyle\int_1^\infty (\mathrm dx)/(x(9+\ln^2(x)))`

or some flavor of this. If this interpretation is correct, substitute u = ln(x) and du = dx/x. Then

`I = \displaystyle\int_0^\infty (\mathrm du)/(9+u^2) \\\\ = \frac13\arctan\left(\frac u3\right)\bigg|_(u=0)^(u\to\infty) \\\\ = \frac13\lim_(u\to\infty)\arctan\left(\frac u3\right) \\\\ = \frac13*\frac\pi2 = \boxed{\frac\pi6}`