Question

More integrals :)

Answer 1

\[\int\limits_{}^{}\frac{ 1 }{ x^2+6x+14 }\] I rewrote this as \[\int\limits_{}^{}\frac{ 1 }{ (x+3)^2+5 }\] but I am not sure that is helpful?

Answer 2

sure it is.
\(\int \dfrac{1}{x^2+a^2}dx = \dfrac{1}{a} arctan x +c\)

Answer 3

but we can't square the 5! unless we write it sqrt5^2

Answer 4

is that legal? o_o

Answer 5

i meant
\(\int \dfrac{1}{x^2+a^2}dx = \dfrac{1}{a} arctan (x/a) +c\)


and yes it is

Answer 6

a = \(\sqrt a^2 \)

Answer 7

wooah
\[\frac{ 1 }{ \sqrt5 }\tan^{-1} (\frac{ x+3 }{ \sqrt5 })+C\]

Answer 8

\(\Huge \checkmark \)

Answer 9

that's it? that's the end? :o

Answer 10

absolutely

Answer 11

when i do
\(\checkmark\)
assume its the end :P

Answer 12

xD ik ik but it was so short haha

Answer 13

completing the square was the majority of it, which you had already done :)

Answer 14

Well yay! :>
I guess we could rewrite this
\[\frac{ \sqrt5 }{ 5 }\tan^{-1}(\frac{ \sqrt5(x+3) }{ 5 }+C\] but that doesn't make much of a difference xD

Answer 15

it does,
it makes the answer look more ugly :P

Answer 16

*sniffle. fiiine.

Answer 17

onto next one! :)