Question

@SolomonZelman
here is an integral for you lol
\(\large \int \frac{2x^2+2x-6}{x-6}dx\)
if you still like them hehe

Answer 1

Hey how a bout some magic zeros?
\[\Large \int\limits_{ }^{ } \frac{2x^2-2x-6}{x-6}~dx\]\[\Large 2 \int\limits_{ }^{ } \frac{x^2-x-3}{x-6}~dx\]\[\Large 2 \int\limits_{ }^{ } \frac{x^2-x-3-27+27}{x-6}~dx\]
\[\Large 2 \int\limits_{ }^{ } \frac{x^2-x-30+27}{x-6}~dx\]\[\Large 2 \int\limits_{ }^{ } \frac{x^2-x-30}{x-6}+\frac{27}{x-6}~dx\]

Answer 2

:PPPPPPP

Answer 3

ha ha and factor the top and on...

Answer 4

hmm interesting

Answer 5

i did it differently i first separated it into 3 integrals
then i used u sub u=x-6

Answer 6

The main thing is that it's easy, not that's it's interesting, although it's certainly is:)

Answer 7

my way is easier :PP lol

Answer 8

i guess so^_^

Answer 9

i attempted to do it that way but i didn't see -30 lol

Answer 10

well, i did this on my paper.
What gets me x^2 and -x for my middle term with (x-6) ?
it was (x+5)
....
(x+5)(x-6) the last term is -30.

Answer 11

Oh i see!

Answer 12

it's actually a super helpful thing for lazy ppl.

Answer 13

I'm sometimes lazy lol

Answer 14

and I'm always. well, i mean why make it harder, if it easy in reality?

Answer 15

tnx for throwing a question though, it was good.

Answer 16

lol! my pleasure ^_^

Answer 17

You really did 3 integrals?

Answer 18

So you had to integrate \(x^2/(x-6)\)

Answer 19

I didn't have to? why?

Answer 20

Nothing, I'm just thinking. I guess in the end the \(u=x-6\) is most straight forward approach

Answer 21

u would do u, lol, for me it seemed easy to complete the fraction (using it similarly to "completing the square")

Answer 22

@wio yes that's what i did
and one of the integrals is x^2/x-6 with u sub
but i think @SolomonZelman is easier

Answer 23

' algebra ' is easier than calc, I would assume so:)

Answer 24

^ this guy xD