How can one evaluate $\int_0^1 \frac{3x^3-x^2+2x-4}{\sqrt{x^2-3x+2}}\mathrm dx$ using only elementary means? How can one evaluate $$\int_0^1 \frac{3x^3-x^2+2x-4}{\sqrt{x^2-3x+2}}\mathrm dx$$ using only elementary means?
By using only elementary means I mean methods not exceeding the basic methods taught in a first course in the integral calculus. This is because I want an explanation accessible to secondary school students, as it was asked of me from a student, and I haven't so far come up with a simple way to evaluate this.
Thank you.
 A: Hint: $(x^2-3x+2)'=2x-3$
now  write $$3x^3-x^2+2x-4=l(2x-3)(x^2-3x+2)+m(x^2-3x+2)+n(2x-3)+r$$ by comparing coefficents.....
A: One way:
Let $3x^2-x^2+x-4=A(x^2-3x-2)(2x-3)+B(x^2-3x+4)+C(2x-3)+D.$
Find $A,B,C,D$ by comparing the coefficients of various powers of $x$. then
$$\int_0^1 \frac{3x^3-x^2+2x-4}{\sqrt{x^2-3x+2}}dx=A\int_{0}^{1}\sqrt{x^2-3x+2} ~(2x-3)~dx+B\int_{0}^{1} \sqrt{x^2-3x+2} ~dx+ C\int_{0}^{1}\frac{(2x-3)dx}{\sqrt{x^2-3x+2}}+D\int_{0}^{1} \frac{dx}{\sqrt{x^2-3x+2}}.$$
Solve first and third  integrals by letting $2x-3=t$, second and fourth integra;ls are standard.
A: Hint:
We make the integrand clearer by depleting the denominator with the change of variable $2x-3=t$.
The integral becomes
$$\int_{-3}^{-1}\frac{3t^3+25t^2+77t+55}{8\sqrt{t^2-1}}dt$$ and this calls for the change of variable $\dfrac{dt}{\sqrt{t^2-1}}=-du$, or $t=-\cosh(u)$. (The negative sign is demanded by the integration bounds.)
The term $\cosh^3(u)$ is easy to integrate as $\cosh(u)(\sinh^2(u)+1)$, and $\cosh^2(u)$ as $\dfrac{\cosh(2u)+1}2$. The rest is routine work.
