Интеграл dx/7x(4-ln^2x)

Решите задачу:

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