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.

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 $$\int{\frac{5x}{(2x^2-3) \sqrt{3x^2-2x+1}}}\,\mathrm dx=-\frac{5 \tanh ^{-1}\left(\frac{-\left(2-3 \sqrt{6}\right) x-\sqrt{6}+2}{\sqrt{2 \left(11-2 \sqrt{6}\right)} \sqrt{3    x^2-2 x+1}}\right)}{2 \sqrt{2 \left(11-2 \sqrt{6}\right)}}-\frac{5 \tanh ^{-1}\left(\frac{-\left(2+3 \sqrt{6}\right)    x+\sqrt{6}+2}{\sqrt{2 \left(11+2 \sqrt{6}\right)} \sqrt{3 x^2-2 x+1}}\right)}{2 \sqrt{2 \left(11+2 \sqrt{6}\right)}}$$
