Question

Integration!
\[\int \frac{x^2 + 4}{x + 2} dx\]

Answer 1

Ok, I need a u.
\(u = x^2 + 4\)
\(du = 2x\)
Er.. what do I do with the + 2?
This could get ugly.

Answer 2

nope

Answer 3

\(u = x + 2\) maybe?

Answer 4

long division first

Answer 5

long division works, but I'd complete the square; it's faster

Answer 6

@Euler271 let him do, we gave him the way to do .

Answer 7

\[= \frac{ (x+2)^2 - 4x }{ x+2 } = x+2 - \frac{ 4x }{ x+2 }\]

actually not sure if this is better

Answer 8

Her, actually. :)
And I'm a bit confused at the moment, so... please chime in!

Answer 9

hehehe @Euler271 I found out long division is more beautiful than yours. hehehe

Answer 10

Do you think it'd help any to break it apart?
\[\int \frac{x^2}{x+2} + \frac{4}{x+2} dx\]

Answer 11

nope

Answer 12

\[= x+2 -\frac{ 4(x+2) - 8 }{ x+2 } = x+2 - 4 +\frac{ 8 }{ x+2 }\]

Answer 13

hey, at the end up, you are back to long division, hehehe @Euler271

Answer 14

lol XD true

its arguable which is faster

Answer 15

yes, it 's fun, right?

Answer 16

@Euler271 That looks nicer! Thank you!
\[\int x - 2 + \frac{8}{x+2} dx\]
So that's just \(\frac{x^2}{2} - 2x + \int \frac{8}{x + 2} dx\)
(I'll do that one in a sec)

Answer 17

\[u = x + 2\]\[du = dx\]\[8 \int \frac{1}{u} du = 8\ln |u| + C\]
So...
\[\frac{x^2}{2} - 2x + \ln |x+2| + C\]

Answer 18

Oops, with the 8.

Answer 19

yes

Answer 20

Thank you!