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!
On
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.
On
The Euler substitution $\sqrt{9x^2-9x+2} =t+3x$ is better suited here, resulting in $x=\frac{2-t^2}{3(2t+3)} $, $dx=-\frac23 \frac{t^2+3t+2}{(2t+3)^2}dt$ \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\bigg|\frac{4t+9-\sqrt{17}}{4t+9+\sqrt{17}} \bigg| \end{align}
Try this.
\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}