Question

What did I do wrong here? I am getting ln|1-1|....
http://i1084.photobucket.com/albums/j409/QRAWarrior/MATA36-IMG007a.png

http://i1084.photobucket.com/albums/j409/QRAWarrior/MATA36-IMG007b.png

Answer 1

Its quite fade

Answer 2

try zooming in

Answer 3

Where are people getting this bicycle avatars from?

Answer 4

these*

Answer 5

google

Answer 6

Is it \[\int\limits \left(\ln \left(x+\sqrt{x^2+1}\right)-10\right) \, dx\]

Answer 7

The integral is:

The definite integral from 1 to \sqrt{2} of:

ln[x + \sqrt{x^2-1}] dx

Answer 8

did you try wolfram?

Answer 9

No, and I do not want to. That just gives it away, or it does it in some other wacky method

Answer 10

i think it gives the best solution , because other than integration by parts, there's no other way you could do this

Answer 11

\[\int\limits_{?}^{?}\ln (x+\sqrt{x^{2}-1}) dx\]
Let x = sec (w)
dx = sec(w)tan(w) dw
\[\rightarrow \int\limits_{?}^{?} \ln(\sec w + \tan w)* (\sec w \tan w) dw\]
Note:
\[\int\limits_{?}^{?} \sec w = \ln(\sec w +\tan w)\]
\[\frac{d}{dw} \sec w = \sec w \tan w\]
use integration by parts:
u = ln(sec(w) +tan(w)) ........... dv = sec(w) tan(w)

du = sec(w) .......................... v = sec(w)
\[\rightarrow \sec w \ln(\sec w +\tan w) -\int\limits_{?}^{?} \sec^{2} w \]
\[\rightarrow \sec w \ln(\sec w +\tan w) - \tan w\]
substituting back in for x
\[\rightarrow x \ln(x+\sqrt{x^{2}-1}) - \sqrt{x^{2}-1}\]
Now evaluate from 1 to sqrt2

Answer 12

Ok I see, so you are going to use trig sub for the argument. Thanks for the suggestion,

Answer 13

no problem..you don't have to, you can go straight to integration by parts as wolfram does but this way it may be easier if you comfortable with trig

Answer 14

Did you use ( lnu )' ?

Answer 15

what are you referring to ?

Answer 16

Let u = x + sqrt ( x^2 + 1)
du = dx/ sqrt ( x^2 +1)

Answer 17

dv = dx -> v = x

Answer 18

correct...yes except i did a trig substitution first

Answer 19

Integral by part:
= xln( x + sqrt ( x^2 +1) - sqrt ( x^2 -1)

Answer 20

I'm unfamiliar with plugging the values :/