Question

integrate dx/(x+2)square root x^2+4x+8

Answer 1

\[\int\limits_{}^{}\frac{ dx }{ x+2 }\sqrt{x^2+4x+8}\]

Right ?

Answer 2

square root at the bottom

Answer 3

i"ll try to do partial fraction but i got o for a

Answer 4

Use substitution :)

Answer 5

this is my working.,but it does not make sense.,
\[\int\limits_{}^{} \frac{ dx}{ (x+2)(\sqrt{(x+2)^{2}} }\]

Answer 6

is it correct if i just cancel out the squaare root

Answer 7

use (x+2)/2 = tan y

Answer 8

\[\int\limits_{}^{}\frac{ 1 }{ x+2\sqrt{x^2+4x+8} }=\int\limits_{}^{}\frac{ 1 }{ x+2\sqrt{(x+2)^2 +4} }\]

Substiute u=x+2 du=dx

Answer 9

Sorry i missed the dx :/

Answer 10

(x+2)^2 is not equal to x^2+4x+8

Answer 11

I wrote (x+2)^2+4=x^2+4x+8

Answer 12

i said that to gee..

Answer 13

\[\int\limits_{}^{}\frac{ 1 }{ u+\sqrt{u^2+4 } }du\]

Now substitute

u=2tan(y) and du=2sec^2 ydy

Answer 14

@hartnn ouh ok.,i did not noticce it is polar

Answer 15

@hba
its multiplication in the denominator there..

Answer 16

\(\int\limits_{}^{} \frac{ dx}{ 2[(x+2)/2](\sqrt{((x+2)/2)^{2}+1} }\)
now try tan y = (x+2)/2

Answer 17

you should get something of the form, c cosec y

Answer 18

1/usqrt(u^2 +4)
=1/(u^2) sqrt( 1 + 4/u^2)

let 1/u = t

no need for trig substitution if your not comfortable with it..

Answer 19

it then reduces to standard form..

Answer 20

Useful:

\[\csc \theta *\frac{ \csc \theta + \cot \theta }{ \csc \theta + \cot \theta }=\frac{ \csc^2 \theta + \cot \theta \csc \theta }{ \cot \theta + \csc \theta }\]

Answer 21

@hba how u get u=2tan(y) and du=2sec^2 ydy

Answer 22

@hartnn could u continue @hba style?

Answer 23

@hartnn i dont really get ur working

Answer 24

his work is not exactly correct..

Answer 25

ok,
\(\int\limits_{}^{}\frac{ 1 }{( u+2)\sqrt{u^2+4 } }du\)
u substitute here
u=2tan y
du=... ?

Answer 26

whats the problem with my solution? :D

Answer 27

@hartnn ouh really?that"s y i stuck here
\[\int\limits_{}^{} \frac{ du }{ u \sqrt{u^2+4} }\]

Answer 28

oh, sorry
thats is correct

Answer 29

still u put, u=2tan y, what u get ?

Answer 30

du=....?

Answer 31

@shubhamsrg I mean, your way just seems like it doesn't really do anything. It's still a trig substitution after you get it to about here:

\[-\int\limits_{}^{}\frac{ dt }{ \sqrt{1+4t^2} }\]

Answer 32

@shubhamsrg could you please type your work properly.,i could get it

Answer 33

du = 2 sec^2 y dy
\(\large \int \frac{2sec^2 ydy}{2 tan y.4.(secy)}=.....?\)

Answer 34

i think there will be 2 sec y instead of 4 sec y...

Answer 35

@shubhamsrg could not get it

Answer 36

after taking all constants aside, we get main part to integrate is
1/sqrt[ (1/2)^2 + t^2 ]

which is a standard case @Kainui

Answer 37

u got this after substitution ?
\(\large \int \frac{2sec^2 ydy}{2 tan y.2.(secy)}=.....?\)
?

Answer 38

@hartnn where u get u=2tan y?i cant figure it out?

Answer 39

there was u in the denominator....

Answer 40

I guess I just don't see going through all that substitution as being easier, even though the result looks prettier.

Answer 41

@gee