Oh yeah, even if Fire Mage doesn't like The Clash now, that doesn't mean he can't ever like them. I think almost anyone can like them. There are so many genres and bands I originally did not enjoy but I grew to like them through repeated listens or other bands which were closer to my own taste but were also similar to the kind of music I wasn't accustomed to. But even though you can grow to like something you initially didn't, that doesn't mean it's going to be case with everything. Some music just doesn't click with some people. And for example, I think my musical taste is really broad and I listen to almost anything but even so, I still have my favorite bands and maybe even favorite genres, although I kinda have no idea what my favorite genre is right now. But for instance, I definitely listen to modern rock, electronica or metal more than I listen to, say, hip-hop, jazz or punk. I would be lying if I said I liked jazz or punk as much as rock. But I do listen to all that, some in smaller doses but I think there are artists I really like in almost all genres.

I started to a bit off-topic there but basically my point was that even though you have some genres or artists you might prefer over others, you can definitely like any kind of music because there's good music in all genres. So basically what idiotkid said. And besides, I think it's actually really interesting to keep listening different kind of artists even if it's completely different from what you usually listen to and just keep doing and discovering new muzaks then a year later you can think back how much your taste has changed and what kind of stuff you have started to get into.

∫{0..1} (x-1)/(x+1) dx
Split into two integrals:
= ∫{0..1} x/(x+1) dx - ∫{0..1} 1/(x+1) dx

First integral:
∫{0..1} x/(x+1) dx
Let u = x+1 → du = dx and x = u-1
= ∫{x=0..1} (u-1)/u du
= ∫{x=0..1} 1 - 1/u du
Split into two integrals:
= ∫{x=0..1} 1 du - ∫{x=0..1} 1/u du
= u - ln|u| {x=0..1}
= (x + 1) - ln(x + 1) {x = 0..1}
= [(1 + 1) - ln(1 + 1)] - [(0 + 1) - ln(0 + 1)]
= 2 - ln(2) - 1 + 0
= 1 - ln(2)

Second integral:
∫{0..1} 1/(x+1) dx
Let u = x + 1 → du = dx
= ∫{x=0..1} 1/u du
= ln|u| {x=0..1}
= ln|x + 1| {x=0..1}
= ln|1 + 1| - ln|0 + 1|
= ln(2) - ln(1)
= ln(2)

Combining the two integrals:
= [1 - ln(2)] - ln(2)
= 1 - 2ln(2)
= 1 - ln(2^2)
= 1 - ln(4)

That should be right, if not then it can't be too far off!