How to integrate the product of two or more polynomials raised to some powers, not necessarily integral

Solution 1:

Looking at your previous deleted post, one answer suggested to use Euler subtitution $$\sqrt{x^2-3x+2}=t+x\implies x=\frac{2-t^2}{2t+3}\implies dx=-\frac{2 (t+1) (t+2)}{(2 t+3)^2}\,dt$$ Replacing, we arrive to $$\frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 - 3x + 2}}=\frac{2 (t+1)^2 \left(3 t^4-4 t^3-2 t^2+56 t+60\right)}{(2 t+3)^4}$$ Now let $2t+3=u$ to make the integrand $$\frac{3 u^2}{64}-\frac{25 u}{32}+\frac{317}{64}-\frac{135}{16 u}+\frac{317}{64 u^2}-\frac{25}{32 u^3}+\frac{3}{64 u^4}$$ and the antiderivative $$\frac{u^3}{64}-\frac{25 u^2}{64}+\frac{317 u}{64}-\frac{135}{16} \log \left({u}\right)-\frac{317}{64 u}+\frac{25}{64 u^2}-\frac{1}{64 u^3}$$ For $t$, the bounds were $(\sqrt 2,-1)$; so, for $u$, they are $(2\sqrt 2+3,1)$ giving as a result $$ \int_{0}^{1} \frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 - 3x + 2}} \, dx=\frac{135}{16} \log \left(3+2 \sqrt{2}\right)-\frac{101}{4 \sqrt{2}}\approx -2.98127$$

Solution 2:

Alternative method: $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $

Express the integrand in the form $\lfrac{(2ax+b)·(x^2-3x+2)+(2cx+d)}{\sqrt{x^2-3x+2}}$ for some constants $a,b,c,d$.

Then split it into $( a(2x-3) + (3a+b) ) · \sqrt{x^2-3x+2} + \lfrac{c(2x-3)+(3c+d)}{\sqrt{x^2-3x+2}}$, so that as a sum of four terms the first and third have obvious antiderivatives. The other two terms can be solved by standard techniques.