Question

finding integral with hyperbolic function,

Answer 1

\[\int\limits_{}^{} \frac{ 1 }{ \sqrt{x^{2} + 6x + } } dx\]

Answer 2

ah it is x^2 + 6x +8

Answer 3

i factor the polynomial in the sqrt to (x+3)^3 -1,
then i use
\[\sinh^{2} y = (x+3)^{2}\]

Answer 4

dx = cosh y dy

Answer 5

i get Arcsinh(x+3) as my final answer, but it doesn't seem quite right

Answer 6

Sinh = a bunch of exponentials.
So the inverse will involve logarithms.

The correct answer is supposed to be a big logarithm. So you're probably on the right track.
We just need a way to rewrite it. hmm

Answer 7

wait, i think i got something wrong, (x+3)^2 should be (cosh y)^2 cuz i can turn the polynomial to (sinh y)^2

Answer 8

if tat true i should get Arccosh(x+3), after that i need to express them in exponential?

Answer 9

Mmm maybe I missed something somewhere, I thought you were on the right track the first time.
Does the integral simplify to,\[\Large \int\limits dy \quad=\quad y \qquad\to\qquad y=arcsinh(x+3)\]

Answer 10

Oh the identity doesn't work that way does it XD haha my bad!

Answer 11

\[\Large \int\limits \frac{1}{\sqrt{(x+3)^2-1}}dx\]\[\Large x+3=\cosh y \qquad\to\qquad dx=\sinh y\;dy\]\[\Large \int\limits \frac{1}{\sqrt{\cosh^2y-1}}(\sinh y\;dy)\]

Answer 12

Arccosh? ok ok ok i see what you were doing there :3

Answer 13

right? +_+ or no? lol

Answer 14

yeah arc cosh(x+3)

Answer 15

hmm

Answer 16

Well if you're allowed to use an identity, we can use this:\[\Large \cosh^{-1}x \quad=\quad \ln\left(x+\sqrt{x^2+1}\right)\]
But if we need to actually work it out, I'm trying to remember how to get there. :p heh

Answer 17

yeah i am figuring how to eleminate the arccosh

Answer 18

Seems to me like everything you guys did is correct, one would end up with a property of the angle only, or the inverse of the hyperbolic function respective. \[\Large y(x)=\sinh(x)=\frac{1}{2}(e^{-x}+e^x) \]such that:\[\Large x=\frac{1}{2}(e^{-y}+e^y) \]
or after a substitution u=e^y you might get:
\[\Large x=\frac{1}{2}(u^{-1}+u) \]
multiply by u to get the quadratic:
\[\Large 2u^2-2xu+1=0 \]

Answer 19

correction**\[\Large u^2-2xu+1=0 \]
for u=e^y

Answer 20

\[\cosh y +\sinh + y = e^{y} , \cosh y = x+3\]

Answer 21

\[\sinh y = \sqrt{\cosh^{2}-1} = \sqrt{x^{2}+6x + 8}\]

Answer 22

then i get the same answer in wolframalpha, y = ln( x+3 + sqrt(x2 + 6x + 8))

Answer 23

thanks guys