Integral $\int{\frac{5x}{(2x^2-3) \sqrt{3x^2-2x+1}}}dx$ I need to solve the following problem:
$$\int{\frac{5x}{(2x^2-3) \sqrt{3x^2-2x+1}}}dx$$
I tried with trigonometric substitution, but I couldn't solve it. The square root is messing me up. I searched it up and nothing appears to work well so I hope someone can help. Thanks.
 A: This is going to get ugly real quick.
$$\int{\frac{5x}{\left(2x^2-3\right)\sqrt{3x^2-2x+1}}}\,\mathrm dx\equiv 5\sqrt3\int\frac x{\left(2x^2 - 3\right)\sqrt{(3x - 1)^2 + 2}}\,\mathrm dx$$
Let $u = 3x - 1\implies\mathrm dx = \dfrac13\mathrm du$ and $x = \dfrac{u + 1}3, x^2 = \dfrac{(u + 1)^2}9$. Therefore,
$$\int\frac x{\left(2x^2 - 3\right)\sqrt{(3x - 1)^2 + 2}}\,\mathrm dx\equiv\int\dfrac{u + 1}{\left(2(u^2 + 2u) - 25\right)\sqrt{u^2 + 2}}\,\mathrm du$$
Next, substitute $u = \sqrt2\tan(v)\implies\mathrm du = \sqrt2\sec^2(v)\,\mathrm dv$. So,
$$\begin{align}\int\dfrac{u + 1}{\left(2(u^2 + 2u) - 25\right)\sqrt{u^2 + 2}}\,\mathrm du&\equiv\int\dfrac{\sqrt2\sec^2(v)\left(\sqrt2\tan(v) + 1\right)}{\left(2\left(2\tan^2(v) + \sqrt{2^3}\tan(v)\right) - 25\right)\sqrt{2\tan^2(v) + 2}}\,\mathrm dv \\ &\stackrel{\sec^2(v) = 1 + \tan^2(v)}=\int\dfrac{\sec(v)\left(\sqrt2\tan(v) + 1\right)}{2\left(2\tan^2(v) + \sqrt{2^3}\tan(v)\right) - 25}\,\mathrm dv\end{align}$$
Perform tangent half-angle substitution to get
$$\int\dfrac{\sec(v)\left(\sqrt2\tan(v) + 1\right)}{2\left(2\tan^2(v) + \sqrt{2^3}\tan(v)\right) - 25}\,\mathrm dv\equiv\int\dfrac{1 + \left(\tan^2\left(\frac v2\right)\right)\left(\frac{\sqrt{2^3}\tan\left(\frac v2\right)}{1 - \tan^2\left(\frac v2\right)}\right) + 1}{\left(1 - \tan^2\left(\frac v2\right)\right)\left(2\left(\frac{\sqrt{2^5}\tan\left(\frac v2\right)}{1 - \tan^2\left(\frac v2\right)} + \frac{8\tan^2\left(\frac v2\right)}{\left(1 - \tan^2\left(\frac v2\right)\right)^2}\right) - 25\right)}\,\mathrm dv$$
Finally, let $w = \tan\left(\dfrac v2\right)\stackrel{\sec^2(v) = 1 + \tan^2(v)}\implies\mathrm dv = \dfrac2{1 + w}\,\mathrm dw$. Therefore,
$$\int\dfrac{1 + \left(\tan^2\left(\frac v2\right)\right)\left(\frac{\sqrt{2^3}\tan\left(\frac v2\right)}{1 - \tan^2\left(\frac v2\right)}\right) + 1}{\left(1 - \tan^2\left(\frac v2\right)\right)\left(2\left(\frac{\sqrt{2^5}\tan\left(\frac v2\right)}{1 - \tan^2\left(\frac v2\right)} + \frac{8\tan^2\left(\frac v2\right)}{\left(1 - \tan^2\left(\frac v2\right)\right)^2}\right) - 25\right)}\,\mathrm dv\equiv\int\dfrac{2\left(w^2 - \sqrt{2^3}w - 1\right)}{25w^4 + \sqrt{2^7}w^3 - 66w^2 - \sqrt{2^7}w + 25}\,\mathrm dw$$
From here on, you need to factorize the denominator and obtain partial fractions.
A: Integrate as follows\begin{align}
& \int{\frac{x}{(2x^2-3) \sqrt{3x^2-2x+1}}}\ dx\\
=& \ \frac12\int \left(\frac{1}{2 x+\sqrt{6}}+\frac{1}{2 x-\sqrt{6}}\right)\frac1{\sqrt{3
   x^2-2 x+1} }\ dx\>\>\>\>\>\>\>t_{\pm}=\frac1{2x\pm \sqrt6}\\
=& -\frac14\int \frac {1}{\sqrt{3-2(2+3\sqrt6)t_+ +2(11+2\sqrt6)t_+^2}}\ dt_+ \\
&\>-\frac14\int \frac {1}{\sqrt{3-2(2-3\sqrt6)t_- +2(11-2\sqrt6)t_-^2}}\ dt_-\\\
=& \ - \frac{\sinh^{-1}\frac{2(11+2\sqrt6)t_+-(2+3\sqrt6)}{2\sqrt2}}{4\sqrt{2(11+2\sqrt6)}}- \frac{\sinh^{-1}\frac{2(11-2\sqrt6)t_+-(2-3\sqrt6)}{2\sqrt2}}{4\sqrt{2(11-2\sqrt6)}}+C\\
= & \ \frac{\sinh^{-1}\frac{(\sqrt2+3\sqrt3)x -(\sqrt2+\sqrt3)}{2x+\sqrt6}}{4\sqrt{2(11+2\sqrt6)}}+\frac{\sinh^{-1}\frac{(\sqrt2-3\sqrt3)x-(\sqrt2-\sqrt3)}{2x-\sqrt6}}{4\sqrt{2(11-2\sqrt6)}}+C 
\end{align}
A: Disclaimer. In all general cases below, you need to choose your (real) constants such that all terms are well-defined.
First note $2x^2-3=\frac12(2x-\sqrt6)(2x+\sqrt6)$ so we can decompose the integrand as $$\frac{5x}{(2x^2-3) \sqrt{3x^2-2x+1}}=\frac52\cdot\left(\frac{1}{\left(2 x-\sqrt{6}\right) \sqrt{3 x^2-2 x+1}}+\frac{1}{\left(2 x+\sqrt{6}\right) \sqrt{3
   x^2-2 x+1}}\right).$$ For an integral of the type $$\int \frac{1}{(d+ex)\sqrt{a+bx+cx^2}}\,\mathrm dx$$ we can use the substitution $u=\frac{2 ae-bd-(2cd-be)x}{\sqrt{a+bx+cx^2}}$, leading to $$\int \frac{1}{\underbrace{4cd^2-4bde+4ae^2}_{\text{constant}}-u^2}\,\mathrm du.$$ But this integral is well-known as $$\frac{\tanh ^{-1}\left(\frac{u}{2 \sqrt{a e^2-b d e+c d^2}}\right)}{2 \sqrt{a e^2-b d e+c d^2}}$$ and using the definition of $u$ we get $$\int \frac{1}{(d+ex)\sqrt{a+bx+cx^2}}\,\mathrm dx=-\frac{\tanh ^{-1}\left(\frac{2 a e-b d+b e x-2 c d x}{2 \sqrt{a+x (b+c x)} \sqrt{e (a e-b d)+c
   d^2}}\right)}{\sqrt{e (a e-b d)+c d^2}}.$$
In our special case, this leads to $$\int{\frac{5x}{(2x^2-3) \sqrt{3x^2-2x+1}}}\,\mathrm dx=-\frac{5 \tanh ^{-1}\left(\frac{-\left(2-3 \sqrt{6}\right) x-\sqrt{6}+2}{\sqrt{2 \left(11-2 \sqrt{6}\right)} \sqrt{3    x^2-2 x+1}}\right)}{2 \sqrt{2 \left(11-2 \sqrt{6}\right)}}-\frac{5 \tanh ^{-1}\left(\frac{-\left(2+3 \sqrt{6}\right)    x+\sqrt{6}+2}{\sqrt{2 \left(11+2 \sqrt{6}\right)} \sqrt{3 x^2-2 x+1}}\right)}{2 \sqrt{2 \left(11+2 \sqrt{6}\right)}}$$
A: This Is my work:

then solve integral with respect to $u$ and so on
