Problem about evaluating $\int_0^1 {3x^3 -x^2 +2x -4\over \sqrt {x^2-3x+2} } \; dx $ $$\int_0^1 {3x^3 -x^2 +2x -4\over \sqrt {x^2-3x+2} } \; dx $$
I tried to solve it but I couldn't get the right procedure..
I need help how should I start.
 A: Let $f:\Bbb R\setminus\{1, 2\}\to \Bbb R$ denote the integrand. We have
$$f(x)=\frac{3x^3-x^2+2x-4}{\sqrt{(x-\frac32)^2-\frac14}}.$$
By substituting $x_{\text{new}}=x_{\text{old}}-\frac32$, we have (looking at indefinite integrals)
$$\int f(x)\,\mathrm dx = \int \frac{3(x+\frac32)^3-(x+\frac32)^2+2(x+\frac32)-4}{\sqrt{x^2-\frac14}} \,\mathrm dx.$$
Now we can use standard trigonometric substitution to deal with the square root in the denominator:
Substitute $u=\operatorname{arcsec}(2x)$. Then $$\sqrt{x^2-\frac14}=\frac{\tan(u)}2$$ and $$x=\frac{\sec u}2$$ so $$\mathrm dx \sim \frac12 \tan(u)\sec(u)\,\mathrm du.$$
Using this in our integral and doing some calculations we see that $$\int f(x) \,\mathrm dx = \frac18\int3 \sec^4(u)+ 25 \sec^3(u)+77\sec^2(u)+55\sec(u)\,\mathrm du.$$
Can you finish from here? (Hint: Use this)
EDIT: For the improper definite integral I got $$-\frac1{16}\cdot\left(202\sqrt 2+ 135 \ln(3-2\sqrt 2)\right).$$
A: Hint: try with substitution $$\sqrt{x^2-3x+2}=x+t$$ than explicate $x$ in function of $t$ and obtain a fraction integral.
