5) $int\frac{2 + ln^{2}x}{x(1 - ln x)}dx$\n6) $int\frac{sin3t}{cos3t - 1}dt$

Explanation:

Step1: For the first integral $\int\frac{2 + \ln^{2}x}{x(1-\ln x)}dx$

Let $u=\ln x$, then $du=\frac{1}{x}dx$. The integral becomes $\int\frac{2 + u^{2}}{1 - u}du$. We perform polynomial long - division: $\frac{2 + u^{2}}{1 - u}=-u - 1+\frac{3}{1 - u}$. Then $\int\frac{2 + u^{2}}{1 - u}du=-\int udu-\int 1du + 3\int\frac{1}{1 - u}du=-\frac{u^{2}}{2}-u-3\ln|1 - u|+C$. Substituting back $u = \ln x$, we get $-\frac{\ln^{2}x}{2}-\ln x-3\ln|1-\ln x|+C$.

Step2: For the second integral $\int\frac{\sin3t}{\cos3t - 1}dt$

Let $v=\cos3t-1$, then $dv=-3\sin3t\ dt$. So $\int\frac{\sin3t}{\cos3t - 1}dt=-\frac{1}{3}\int\frac{dv}{v}=-\frac{1}{3}\ln|v|+C=-\frac{1}{3}\ln|\cos3t - 1|+C$.

Answer:

1. $-\frac{\ln^{2}x}{2}-\ln x-3\ln|1-\ln x|+C$
2. $-\frac{1}{3}\ln|\cos3t - 1|+C$