Question

integral sqrt(x^2+2x+2)/x

Answer 1

what it's supposed to look like (for clarification) http://www.wolframalpha.com/input/?i=integral+sqrt%28x^2%2B2x%2B2%29%2Fx

Answer 2

i've tried everything i could think of for values to substitute

Answer 3

yeah, nice arcsinh in there :)

try completing the square under the radical? then you can attempt a trig sub perhaps

Answer 4

i did taht so i got sqrt((x+1)^2+1)/x
but from there what do i do? what do i trig sub for?

Answer 5

well, off hand i know that tan^2+1 = sec^2

Answer 6

right

Answer 7

lets (x+1) = tan(u)

Answer 8

i tried that and then i ended up getting the integral of (secx)^3 / tanx-1 http://www.wolframalpha.com/input/?i=integral+sec^3x%2F%28tanx-1%29

Answer 9

sorry, in this case x should be u

Answer 10

and then from there idk what to do

Answer 11

\[\int \sec(u)\frac{\sec^2(u)}{\tan(u)+1}du\]that at least has ln integral sitting in it; my thoughth is a by part perhaps?

Answer 12

"ln integral" what do you mean by that? can you point that out for me

Answer 13

sec^2 is the derivative of tan+1: u'/u integrates to ln u

Answer 14

so do you think i should do integration by parts?

Answer 15

well, its just sitting there ... might as well give it a shot :)

Answer 16

hm i still dont know how to make use of it

Answer 17

\[\sec(u)~\ln(tan(u)+1)-\int \ln(\tan(u)+1)~\sec(u)\tan(u)~du\]

Answer 18

did you achieve this result via integration by parts?

Answer 19

if so, what were your terms?

Answer 20

well, since the integration was already set; i used sec^2/(tan+1) as dv, and sec as u

Answer 21

not sure of that int v du is workable tho ....

Answer 22

hm ok i understand the integration by parts. let me try to work out the second integral now haha

Answer 23

.... im pretty sure its messier than the first :)

Answer 24

yeah lol

Answer 25

well through a bunch of other various sdbustitutions

Answer 26

i arrived at the integral of sqrt(u^2+1)/(u-1) http://www.wolframalpha.com/input/?i=integral+of+sqrt%28u^2%2B1%29%2F%28u-1%29

Answer 27

i wonder if there some multiplier of 1 that would make that sqrt vanish

Answer 28

if we can solve that simpler one then we solve the original question

Answer 29

try working on the new one to see. i got stuck on that one also

Answer 30

\[\frac{\sqrt{x^2+2x+2}}{x}\]
\[\frac{x^2+2x+2}{x\sqrt{x^2+2x+2}}\]
\[\frac{x+2}{\sqrt{x^2+2x+2}}+\frac{2}{x\sqrt{x^2+2x+2}}\]
\[\frac22\frac{x+2}{\sqrt{x^2+2x+2}}+\frac{2}{x\sqrt{x^2+2x+2}}\]

Answer 31

the left part is simple enough, thats another ln up; the right part is looking triggy now

Answer 32

WOAH

Answer 33

lol, not ln but sqrt

Answer 34

ok so the integral of 1/sqrt(x^2+1) is that sinhx?

Answer 35

i think so, right?

Answer 36

we never really covered the hyperbolics in my class so im not as familiar with them

Answer 37

arsinh***

Answer 38

\[2\sqrt{x^2+2x+2}~+~2\int\frac{1}{x\sqrt{(x+1)^2+1}}dx\]

Answer 39

yeah

Answer 40

if we let u = 1/x and dv = 1/sqrt((x+1)^2+1) dx

Answer 41

u = -1/x^2 and v = inverse sinh(x+1)

Answer 42

du = -1/x^2 dx*

Answer 43

hm well i guess i'll close this question for now. i think simplifying it to that last integral was a lot more help than i was expecting. thank you so much amistre64 :)

Answer 44

youre welcome :)