Question

see below for question

Answer 1

∫ln⁡(x^2+4)ⅆx

Answer 2

interesting box. what is in it?

Answer 3

itz basically the integration of ln(x^2 + 4)

Answer 4

ok parts is what you want

Answer 5

anything as long as itz integetrated

Answer 6

*integrated

Answer 7

get rid of the log by taking
\[u=ln(x^2+4), du=\frac{2x}{x^2+4}\] dv = dx, v = x\]

Answer 8

i got dat part, itz just dat i get stuck in the process

Answer 9

damn preview
\[u=\ln(x^2+4), du = \frac{2x}{x^2+4}, dv = dx, v = x\]

Answer 10

ok so first thing you just write down. it is
\[x\ln(x^2+4)-2\int \frac{x^2}{x^2+4}dx\] and the second one you have to actually either divide, or use a trick

Answer 11

the trick looks snappier, write
\[\frac{x^2}{x^2+4}=\frac{x^2+4-4}{x^2+4}=1-\frac{4}{x^2+4}\] but if you don't think of the trick you divide and get the same thing

Answer 12

THANK YOU!!!! i think i can take it from there!!!!

Answer 13

integral of 1 is x, and integral of
\[\frac{1}{x^2+4}\] is
\[\frac{1}{2}\tan^{-1}(\frac{x}{2})\]and of course you have to do all the adjustments with the constants which i ignored

Answer 14

yw