Решите интеграл: dx/(x-1)sqrt(x^2+3x)

Решите задачу:

$\int \frac{dx}{(x-1)\sqrt{x^2+3x}}=[x-1=\frac{1}{t},\; x=1+\frac{1}{t}\; ,\; dx=-\frac{dt}{t^2}\; ,\\\\x^2+3x=1+\frac{2}{t}+\frac{1}{t^2}+3+\frac{3}{t}= \frac{1}{t^2} +\frac{5}{t}+4= \frac{4t^2+5t+1}{t^2} \; ]=\\\\=-\int \frac{t\cdot dt}{t^2\cdot \frac{1}{t}\sqrt{4t^2+5t+1}} =-\int \frac{dt}{\sqrt{4t^2+5t+1}} =-\int \frac{dt}{2\cdot \sqrt{(t+\frac{5}{8})^2-\frac{9}{64}}} =$

$=-\frac{1}{2}ln\left |t+\frac{5}{8}+\sqrt{(t+\frac{5}{8})^2-\frac{9}{64}}\right |+C=\\\\=-\frac{1}{2}ln\left | \frac{1}{x-1}+\frac{5}{8}+\sqrt{(\frac{1}{x-1}+\frac{5}{8})^2-\frac{9}{64}}\right |+C$