<s>[INST] Finding value of $ \int \frac{1}{x-\sqrt{9x^2+6x+4}}dx$ 
Finding value of $\displaystyle \int \frac{1}{x-\sqrt{9x^2+6x+4}}dx$

Let $$\displaystyle I = \int\frac{1}{x-\sqrt{9x^2+6x+4}}dx = \int\frac{x+\sqrt{9x^2+6x+4}}{-8x^2-6x-4}dx$$
$$I = -\frac{1}{8}\int \frac{x}{x^2+3x/4+1/2}dx-\int\frac{\sqrt{9x^2+6x+4}}{8x^2+6x+4}dx$$
$$I=-\frac{1}{2}\int\frac{x}{\bigg(x+\frac{3}{8}\bigg)^2+\bigg(\frac{\sqrt{33}}{8}\bigg)^2}$$
Now substitute $\displaystyle x+\frac{3}{8}=t$ and $dx=dt$
I am struck for second part of $I$
Could some help me how to solve second part of integral $I$
 [/INST] Hint: Use $9x^2+6x+1=(3x+1)^2$ and let $3x+1=\sqrt{3}\tan\theta$.
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