TITLE: How do you evaluate $\int_{0}^{1} \frac{(3x^3-x^2+2x-4)}{\sqrt{x^2-3x+2}} \, dx$?
QUESTION [6 upvotes]: Saw this problem on a FaceBook meme that said the pin code to his ATM debit card is the solution to the following problem:
$$\int_{0}^{1} \frac{(3x^3-x^2+2x-4)}{\sqrt{x^2-3x+2}} \, dx$$
I was trying to see how we could break this up into an easier integrals but nothing comes to mind at first glance. Perhaps complex integration is possible?

REPLY [2 votes]: Hint:
Substitute $2x-3=-\cosh t$, or $x=\dfrac{3-\cosh t}2$.
We have
$$\int_0^1\frac{3x^3-x^2+2x-4}{\sqrt{x^2-3x+2}}dx=-\int_0^{\text{arcosh }2}(3x^3-x^2+2x-4)\frac{\dfrac{\sinh t}2}{\dfrac{\sinh t}2}dt.$$
Then
$$3x^3-x^2+2x-4=-\frac{-3\cosh^3t+25\cosh^2t-77\cosh t+55}8\\
=-\frac1{32}\cosh 3t+\frac7{16}\cosh 2t-\frac{95}{32}\cosh t+\frac{137}{16}.$$
The rest is routine work.