Question

help please..
using trig sub

Answer 1

\[\int\limits_{}^{}\frac{ dx }{ \sqrt{x^2+2x+5}}\]

Answer 2

@zepdrix

Answer 3

ik its completing the square but did is what i did

\[x^2+2x +4\]
\[x^2 +2x - (1) +5\]

Answer 4

when we complete the square we take half of middle term square it and then add to both sides
\[x^2 +bx +C\]
\[(x+\frac{b}{2})^2 +(\frac{b}{2})^2 = - C + (\frac{b}{2})^2\]
so
\[x^2+2x+\color{Red}{(\frac{2}{2})^2} = -5 \color{Red}{+ (\frac{2}{2})^2}\] \[\color{blue}{x^2+2x +1 }= -5 +1\]
now factor

Answer 5

x^2+bx+C=0 **

Answer 6

\[(x+1)^2=-4\]

Answer 7

yes :)

Answer 8

oh okay so then how would i plug this whole thing back to my original ?

Answer 9

would be like this ?
\[\int\limits_{}^{}\frac{ dx }{ \sqrt{(x+1)^2}}\]

Answer 10

set it equal to 0 \[(x+1)^2 +4 \]
so now you can write x^2+2x+5 as (x+1)^2+4 \[\int\limits_{ }^{ } \frac{1}{\sqrt{(x+1)^2+4}} dx\]

Answer 11

oh okay so then

we are gonna use this one

\[x=a \tan \theta\]

so then

would it be

x+1= 2 tan theta

Answer 12

hmm working on it
i thought we can apply the derivative of sin^{-1} but... it's a^2 +x^2

Answer 13

hmm i got this one sec ima upload it

Answer 14

ohh okay im thinking abt u sub here hmmm

Answer 15

is it blurry or is it okay ?

Answer 16

i can see it ;P

Answer 17

okie lol

Answer 18

hmm that looks good to me
but i'm still confused why did you write (x+1)/2 = tan theta hmm

Answer 19

oh getting tan by itself

Answer 20

oh yeah that's correct \[ \int\limits \frac{ 2 \sec^2 \theta }{ \sqrt{(2\tan \theta )^2 +4 } } = \int\limits_{ }^{} \frac{2 \sec^2 \theta }{\sqrt{4(\tan^2 \theta +1)}}\] \[\int\limits_{ }^{} \frac{ 2 \sec^2 \theta }{ 2 \sec \theta } = \ln \left| \sec \theta + \tan \theta \right|\]
now find sec theta
and tan theta

Answer 21

okie tysm ^_^

Answer 22

yw. good job!! :D