Question

need some help with this integral...

Answer 1

the one in blue?

Answer 2

so the solution manual says to use integration by parts. but when i set up my du i have du=\[du=( 1/(x+\sqrt{x^2 -1}))((1/(2\sqrt{x^2-1}))\]

Answer 3

yeah, how do they simplify to get that?

Answer 4

im using \[u*v-\int\limits_{}^{} v du\]

Answer 5

but im not seeing how they get it down to that

Answer 6

not sure what the question is
the solution is written out
nothing is simplified
is the question
"what is \(\int\frac{x}{\sqrt{x^2-1}}dx\) ?

Answer 7

no, how did they get that?

Answer 8

the integration is of
\[\int \ln(x+\sqrt{x^2-1})dx\]
they chose \(u=\ln(x+\sqrt{x^2-1})\) and \(dv=dx\) making \(v=x\) and \(du=\frac{1}{\sqrt{x^2-1}}dx\)

Answer 9

that makes \(vdu=\frac{x}{\sqrt{x^2-1}}\)

Answer 10

no dude, thats not du.

Answer 11

its;s what i posted above

Answer 12

it is if \(u=\ln(x+\sqrt{x^2-1})\)

Answer 13

chain rule

Answer 14

all the work to get the derivative is in the top line of you post

Answer 15

is that what the question is? the first line is the chain rule, the rest is algebra

Answer 16

maybe the algebra...

Answer 17

ok so first step is clear, i think, it is what you wrote too

Answer 18

second step is to add \[{1+\frac{x}{\sqrt{x^2-1}}}\]

Answer 19

that gives you
\[\frac{\sqrt{x^2-1}+x}{\sqrt{x^2-1}}\]

Answer 20

when you multiply after that, the numerator and denominator cancel leaving \(\frac{1}{\sqrt{x^2-1}}\)

Answer 21

aaaaaaaaaaahh

Answer 22

pellet man, i've been in this library too long..

Answer 23

can't even do algebra

Answer 24

lmoa, i see it now, thanks for the help man