Finding value of $\displaystyle \int \frac{1}{x-\sqrt{9x^2+6x+4}}dx$

Let $$\displaystyle I = \int\frac{1}{x-\sqrt{9x^2+6x+4}}dx = \int\frac{x+\sqrt{9x^2+6x+4}}{-8x^2-6x-4}dx$$

$$I = -\frac{1}{8}\int \frac{x}{x^2+3x/4+1/2}dx-\int\frac{\sqrt{9x^2+6x+4}}{8x^2+6x+4}dx$$


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 $I$



An idea to solve the integral would be this line: $$\int\frac{x}{\bigg(x+\frac{3}{8}\bigg)^2+\bigg(\frac{\sqrt{33}}{8}\bigg)^2}$$ $$\int\frac{x}{\bigg(x+\frac{3}{8}\bigg)^2+ \frac{33}{64}}$$ $$64\int\frac{x}{\bigg(8x + 3\bigg)^2+ 33}$$

Write $x$ as $16 \cdot \frac{1}{128}(8x + 3) −\frac38$ and split:

$$\int \left( \frac{8x + 3}{8\left((8x + 3)^2+33\right)} - \frac{3}{8\left((8x + 3)^2+33\right)}\right) \,dx$$

$$\underbrace{\frac18 \int \frac{8x + 3}{(8x + 3)^2+33}\,dx}_{I_1} - \underbrace{\frac38 \int\frac{1}{(8x + 3)^2+33}}_{I_2}$$

and continue for $I_1 $with

$$u =(8x+3)^2+33 \to dx = \frac1{16(8x+3)}\,du$$

$$I_1=\frac1{16}\int \frac{1}{u}$$

and $I_2$ $$ u=\frac{8x+3}{\sqrt{33}} \to dx=\frac{\sqrt{33}}{8}\,du $$

$$I_2 = \int \frac{\sqrt{33}}{8(33u^2 + 33)} \,du= \frac{1}{8\cdot\sqrt{33}}\int \frac{1}{u^2 + 1}$$


Hint: Use $9x^2+6x+1=(3x+1)^2$ and let $3x+1=\sqrt{3}\tan\theta$.


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