Question

Интеграл dx/7x(4-ln^2x)

Answer 1

$intfrac{dx}{7x(4-ln^2x)}=frac{1}{7}int frac{1}{4-ln^2x} cdot frac{dx}{x} =[, t=lnx,; dt=frac{dx}{x}, ]=\\=frac{1}{7}cdot int frac{1}{4-t^2}cdot dt=-frac{1}{7}cdot int frac{dt}{t^2-4}=\\=-frac{1}{7}cdot frac{1}{2cdot 2}cdot lnleft |frac{t-2}{t+2}right |+C=-frac{1}{28}cdot lnleft | frac{lnx-2}{lnx+2} right |+C$