# Help solving $\int \:\frac{dx}{x+\sqrt{9x^2-9x+2}}$

Reading an article online I came across this integral I've been trying to solve since yesterday morning and I literally tried every integration method I know and I still can't solve it. I am a beginner to integrals so help will be appreciated!

I tried substitution, integration by parts and I can't seem to find the correct way!

$$\int \:\frac{dx}{x+\sqrt{9x^2-9x+2}}$$

Does anyone know how to proceed from here? What to do? Because I am clueless. Thanks!

• For what it's worth you get a closed-form (long) expression when you plug this into WolframAlpha. Jan 27, 2021 at 8:40
• Link to the above comment's reference. Are you sure there isn't a typo in your question? Jan 27, 2021 at 8:44

The Euler substitution $$\sqrt{9x^2-9x+2} =t+3x$$ is better suited here, which results in $$x=\frac{2-t^2}{3(2t+3)}$$, $$dx=-\frac23 \frac{t^2+3t+2}{(2t+3)^2}dt$$ and \begin{align} &\int \:\frac{1}{x+\sqrt{9x^2-9x+2}}dx\\ =& -2 \int \frac{t^2+3t+2}{(2t+3)(2t^2+9t+8)}dt\\ =&-\frac12 \int\left(\frac1{2t+3}+\frac t{2t^2+9t+8} \right)dt\\ =& -\frac14\ln|2t+3| -\frac1{8}\ln| 2t^2+9t+8|+\frac9 {8\sqrt{17}}\ln|\frac{4t+9-\sqrt{17}}{4t+9+\sqrt{17}} |+C \end{align}
Multiply by the conjugate first to make $$\int\frac{dx}{x+\sqrt{9x^2-9x+2}}=\int\frac{\sqrt{9 x^2-9 x+2}-x}{8 x^2-9 x+2} \,dx$$ $$I_2=\int\frac{x}{8 x^2-9 x+2} \,dx=\frac 1{16}\int\frac{16x-9+9}{8 x^2-9 x+2} \,dx$$ $$I_2=\frac 1{16}\int\frac{16x-9}{8 x^2-9 x+2} \,dx+\frac 9{16}\int\frac{dx}{8 x^2-9 x+2}$$ The first one is obvious and the second one is simple (if required, use partial fractions).
Now, we are left with $$I_1=\int\frac{\sqrt{9 x^2-9 x+2}}{8 x^2-9 x+2} \,dx$$ which looks to be much more difficult. However, partial fraction decomposition, we have $$\frac 1{8 x^2-9 x+2}=\frac{1}{8 (a-b) (x-a)}-\frac{1}{8 (a-b) (x-b)}$$ and you can find the required integral in the "Table of Integrals, Series, and Products" (seventh edition) by I.S. Gradshteyn and I.M. Ryzhik.
\begin{align} \int \:\frac{dx}{x+\sqrt{9x^2-9x+2}} &={\displaystyle\int}\left(\dfrac{\sqrt{9x^2+9x+2}}{8x^2+9x+2}-\dfrac{x}{8x^2+9x+2}\right)\mathrm{d}x \\ {\displaystyle\int}\dfrac{\sqrt{9x^2+9x+2}}{8x^2+9x+2}\,\mathrm{d}x &={\displaystyle\int}\dfrac{\sqrt{\left(3x+\frac{3}{2}\right)^2-\frac{1}{4}}}{8x^2+9x+2}\,\mathrm{d}x\tag{solve for 1st integral} \\ &=\frac12 {\displaystyle\int}\dfrac{\sqrt{9\left(2x+1\right)^2-1}}{8x^2+9x+2}\,\mathrm{d}x \\ &=\frac12{\displaystyle\int}\dfrac{\sqrt{9u^2-1}}{4u^2+u-1}\,\mathrm{d}u \tag{u = 2x+1, \mathrm{d}x=\dfrac{1}{2}\,\mathrm{d}u} \\ &=\frac12 {\displaystyle\int}\dfrac{\sqrt{\sec^2\left(v\right)-1}\sec\left(v\right)\tan\left(v\right)}{3\left(\frac{4\sec^2\left(v\right)}{9}+\frac{\sec\left(v\right)}{3}-1\right)}\,\mathrm{d}v \tag{u=\dfrac{\sec\left(v\right)}{3}, v=\operatorname{arcsec}\left(3u\right), \mathrm{d}u=\dfrac{\sec\left(v\right)\tan\left(v\right)}{3}\,\mathrm{d}v} \\ &=\frac32 {\displaystyle\int}\dfrac{\sec\left(v\right)\tan^2\left(v\right)}{4\sec^2\left(v\right)+3\sec\left(v\right)-9}\,\mathrm{d}v \\ &=\frac32 {\displaystyle\int}\dfrac{\frac{\left(\tan^2\left(\frac{v}{2}\right)+1\right) \,\cdot\, 4\tan^2\left(\frac{v}{2}\right)}{\left(1-\tan^2\left(\frac{v}{2}\right)\right)^3}}{\left(\frac{4\left(\tan^2\left(\frac{v}{2}\right)+1\right)^2}{\left(1-\tan^2\left(\frac{v}{2}\right)\right)^2}+\frac{3\left(\tan^2\left(\frac{v}{2}\right)+1\right)}{1-\tan^2\left(\frac{v}{2}\right)}-9\right)}\,\mathrm{d}v \\ & \end{align}