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

Question

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 $I$

Answer 1

HINT:

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

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

Answer 2

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

Answer 3

Complete the square to apply the trigonometric substitution $3x+1=\sqrt3\tan t$, or $$h(x)=\sqrt{9x^2+6x+4}=\sqrt3\sec t$$

\begin{align} &I=\int \frac{1}{x-\sqrt{9x^2+6x+4}}\ dx =\int \frac{\sec t}{\sin t-\frac1{\sqrt3}\cos t-3}\ dt\\ \end{align} Then, decompose the integrand as follows

\begin{align} & \frac{-8\sec t}{\sin t-\frac1{\sqrt3}\cos t-3}\\ =& \ 3\sec t +\frac{\sec t-3\tan t}{\sin t-\frac1{\sqrt3}\cos t-3 } +\frac{\sqrt3}{\sin t-\frac1{\sqrt3}\cos t-3 } \end{align} The three resulting integrals are readily evaluated: $\int \sec t\ dt =\tanh^{-1}\sin t$ \begin{align} & \int\frac{\sec t-3\tan t}{\sin t-\frac1{\sqrt3}\cos t-3 }\ dt = \ln | \tan t-\frac1{\sqrt3}-3 \sec t |\\ & \int \frac{1}{\sin t-\frac1{\sqrt3}\cos t-3 }\ dt = \frac{6}{\sqrt{23}}\tan^{-1}\frac{(1-3\sqrt3)(\csc t-\cot t)+\sqrt3}{\sqrt{23}} \end{align} After back-substitute $$I=-\frac38\tanh^{-1}\frac{3x+1}{h(x)}-\frac18\ln\left(h(x)-x\right)\\ \hspace{30mm}-\frac3{4\sqrt{23}}\tan^{-1}\frac{3\sqrt3 x-(3\sqrt3-1)h(x)+9}{\sqrt{23}\ (3x+1)} $$