Найти неопределенный интеграл (2x-13)/(корень3x^2 - 3x-16)dx

Решите задачу:

$\int \frac{2x-13}{\sqrt{3x^2-3x-16}}dx=[\; 3x^2-3x-16=3(x^2-x)-16=\\\\=3(x-\frac{1}{2})^2-\frac{3}{4}-16=3(x-\frac{1}{2})^2-\frac{67}{4}=3\cdot \Big ((x-\frac{1}{2})^2-\frac{67}{12}\Big ) \; ]=\\\\=\frac{1}{\sqrt3}\cdot \int \frac{2x-13}{\sqrt{(x-\frac{1}{2})^2-\frac{67}{12}}}dx=[\; x- \frac{1}{2}=t\; ,\; x=t+\frac{1}{2}\; ,dx=dt\; ]=\\\\= \frac{1}{\sqrt3} \cdot \int \frac{2t-12}{\sqrt{t^2- \frac{67}{12} }} dt=\frac{1}{\sqrt3}\cdot \int \frac{d(t^2-\frac{67}{12})}{\sqrt{t^2- \frac{67}{12} }}dt -\frac{1}{\sqrt3}\cdot \int \frac{dt}{\sqrt{t^2-\frac{67}{12}}}dt=$

$= \frac{1}{\sqrt3}\cdot 2\sqrt{t^2- \frac{67}{12} } - \frac{1}{\sqrt3} \cdot ln\Big |t+\sqrt{t^2-\frac{67}{12}}\Big |+C=\\\\=\frac{2}{\sqrt3}\cdot \sqrt{(x-\frac{1}{2})^2-\frac{67}{12} }- \frac{1}{\sqrt3}\cdot ln\Big |x-\frac{1}{2}+\sqrt{(x-\frac{1}{2})^2-\frac{67}{12}}\Big |+C$