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

Solution 1:

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}

Solution 2:

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.

The global result will not look very nice.