On a friend's past exam is the problem of integrating $$\int \frac{(x-1)^4}{(x-2)^7}dx.$$ It's pretty easy. Let $u = x-2$. Then $u+1 = x-1$, so we obtain $$\int \frac{(u+1)^4}{u^7}du.$$ Applying the binomial theorem to the numerator and expanding the fraction, we get a Laurent polynomial in $u$ which is straightforwardly integrated.
Now possibly this is a silly question, but I can't help but wonder however if there isn't an easier way of doing this. Are there any tricks for problems of this type that allow us to avoid an appeal to the binomial theorem?
Put $x=2+1/y$, so $dx=-dy/y^2$, and the integral becomes $$ \int \frac{y^7(1+1/y)^4}{y^2} \, dy = \int -y(1+y)^4 \, dy. $$ This can be done with one integration by parts, $$ \int -y(1+y)^4 \, dy = -\frac{1}{5}y(1+y)^5 + \frac{1}{5}\int (1+y)^5 \, dy \\ = -\frac{1}{5}y(1+y)^5 + \frac{1}{30}(1+y)^6 +C. $$ One can do even less calculus (but slightly more algebra) with the substitution $x = 2+\frac{1}{z-1}$, which simply gives $$ \int (z^4-z^5) \, dz. $$