Question

how did they get this answer:

integrate 1/(x^2 + 2x +5)....

Answer 1

yeh what are you typing?

Answer 2

\[= 1/((x+1)^2 + 4\]
u= x+1
du = 1 dx

\[\int\limits_{?}^{?}1/u^2 +4 = 1/2 \tan^{-1} (u/2) +c \]

what? how integrate 1/(x^2 +1) but where is the 1/2 coming from?

Answer 3

it is coming from the 4

Answer 4

the rest of the problem and that bottom integrate is for arctan
\[\int\limits_{?}^{?}1/x^2 +1 = \tan^{-1} x\]

Answer 5

because 4 = 2^2

Answer 6

look at this formula sheet http://www.ecalc.com/math-help/worksheet/calculus-integrals

Answer 7

http://www.ecalc.com/pics/math/calculus_integral.gif

look where it says common integrals, and look at the integral 1/ (u^2 + a^2)

Answer 8

ah im back-- openstudy bonked for a moment --- that equation is a little different than the one in my book...gonna be hard to relearn. And the u's being interchanged ugh. That had to be on purpose.

Answer 9

how is the one in ur book different?

Answer 10

It just has: \[\int\limits_{?}^{?}1/x^2 + 1 = \tan^{-1} x + c\]

Answer 11

yes my bok says integral of du/(u^2 + a^2)=1/a tan^-1 u/a + c

Answer 12

yes, so the answer makes sense knowing that a= 2 in this case

Answer 13

4=a^2 therefore a=2

Answer 14

with the new formula it makes sense...with the old original not so much

Answer 15

lol ok great!