Question

evaluate integral: ln(x + sqrt(x^2 -1)) dx

Answer 1

\[\int\limits \ln (x+\sqrt{x^{2}-1}) dx\]

Answer 2

try by parts

Answer 3

I know you let \[x = \sec^2 \theta\]
\[dx = 2\sec^2 \theta \tan \theta d \theta\]
so \[2 \int\limits \ln(\sec^2 \theta +\tan \theta ) (\sec^2 \theta \tan \theta) d \theta\]

Answer 4

\[\int \ln \spadesuit~dx = x\ln \spadesuit - \int x(\ln \spadesuit )' ~dx \]

Answer 5

\[\int \ln (x+\sqrt{x^{2}-1}) dx =x\ln (x+\sqrt{x^{2}-1}) - \int x \left(\ln (x+\sqrt{x^{2}-1})\right)' dx \]

Answer 6

oh did they cover integration by parts yet ?

Answer 7

we went over integration by parts already, we're on the part where they combined all integration strategies, and we're trying to familiarize which one we use when we see a problem.

Answer 8

good, integration by parts is best for this problem
give it a try

Answer 9

you should think of by parts if you see log term in integrand

Answer 10

did you already start integration by parts for the equation you gave me?
uv - ingetral vdu?

Answer 11

yes

Answer 12

this one is a hard one. did you let u = ln dx and dv = (x + sqrt(x^2-1) ?

Answer 13

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

\(\large v = x\)

Answer 14

\[\large (uv)' = uv' + u'v \implies uv = \int uv' + u'v\]

Answer 15

i was taught to let something be u and other be dv. you derive u to get du, and integral dv to get v.

Answer 16

yes, dv = 1
you get v = x

Answer 17

ah ha....

Answer 18

is there a reason why you let the whole thing equal u. is there something that pops out that made you do that?

Answer 19

or is it trial and error and as you do more practice problem you know what to do.

Answer 20

letting that whole thing equal to u comes naturally once you know how by parts works

Answer 21

\[\large \int \ln (x+\sqrt{x^{2}-1}) dx = \int \ln (x+\sqrt{x^{2}-1}) \color{red}{\cdot 1}~ dx\]

Answer 22

the antiderivative of 1 is x
so you want to integrate 1, doing anything else is ridiculolus here

Answer 23

okay, i got the equation that you posted earlier.

Answer 24

i was just trying to understand why you decided to let u be = to that and dv = to 1

Answer 25

thats a good question actually :)
heard of ILATE mnemonic for by parts ?

Answer 26

no, i have not heard of that.

Answer 27

that helps u in deciding what to pick for dv fast

Answer 28

nope. it helps u in deciding what to pick for `u` fast

Answer 29

http://www.utdallas.edu/dept/abp/PDF_Files/Calculus_Folder/Calculus_I/IntegrationByParts.pdf

Answer 30

so i is first, L is second, A is third etc etc?

Answer 31

oh you let exponential go first, then T, then A?

Answer 32

yes if u see a `log` term and an `exponent` in your integrand,
you pick `log` as `u`

Answer 33

exponential is always the last preference for `u`
because it wont reduce no matter how many times you differentiate

Answer 34

ahhh, okay. let me see what i can do with the resource you gave me.

Answer 35

algebra are x^2 x^3, and exponentials are e^x right?

Answer 36

yes

Answer 37

For your present problem :

\[\begin{align}&\int \ln (x+\sqrt{x^{2}-1}) dx =x\ln (x+\sqrt{x^{2}-1}) - \int x \left(\ln (x+\sqrt{x^{2}-1})\right)' dx\\~\\&= x\ln (x+\sqrt{x^{2}-1}) - \int x \left(\dfrac{1}{x+\sqrt{x^{2}-1}} \right)\left(x+\sqrt{x^{2}-1}\right)' dx\end{align}\]

Answer 38

is the answer \[= xln(x+\sqrt{x^2-1}) - arcsec(x) + C\]

Answer 39

i don't think i derived u correctly.

Answer 40

\[\begin{align}&\int \ln (x+\sqrt{x^{2}-1}) dx =x\ln (x+\sqrt{x^{2}-1}) - \int x \left(\ln (x+\sqrt{x^{2}-1})\right)' dx\\~\\&= x\ln (x+\sqrt{x^{2}-1}) - \int x \left(\dfrac{1}{x+\sqrt{x^{2}-1}} \right)\left(x+\sqrt{x^{2}-1}\right)' dx \\~\\
&= x\ln (x+\sqrt{x^{2}-1}) - \int x \left(\dfrac{1}{x+\sqrt{x^{2}-1}} \right)\left(1+\dfrac{2x}{2\sqrt{x^{2}-1}}\right) dx

\end{align}\]

Answer 41

okay i got that.

Answer 42

\[\begin{align}&\int \ln (x+\sqrt{x^{2}-1}) dx =x\ln (x+\sqrt{x^{2}-1}) - \int x \left(\ln (x+\sqrt{x^{2}-1})\right)' dx\\~\\&= x\ln (x+\sqrt{x^{2}-1}) - \int x \left(\dfrac{1}{x+\sqrt{x^{2}-1}} \right)\left(x+\sqrt{x^{2}-1}\right)' dx \\~\\

&= x\ln (x+\sqrt{x^{2}-1}) - \int x \left(\dfrac{1}{x+\sqrt{x^{2}-1}} \right)\left(1+\dfrac{2x}{2\sqrt{x^{2}-1}}\right) dx\\~\\
&= x\ln (x+\sqrt{x^{2}-1}) - \int x \left(\dfrac{1}{x+\sqrt{x^{2}-1}} \right)\left(\dfrac{\sqrt{x^{2}-1}+x}{\sqrt{x^{2}-1}}\right) dx\\~\\

&= x\ln (x+\sqrt{x^{2}-1}) - \int\left( \dfrac{x}{\sqrt{x^{2}-1}}\right) dx\\~\\

\end{align}\]

Answer 43

substitute `u^2 = x^2-1` to evaluate the integral

Answer 44

is there a reason why you did u^2 = x^2 -1 instead of u = x^2 -1

Answer 45

is it because its under a sq rt?

Answer 46

Yooooon!
Your original approach was just fine.
You were just off by a little bit with your substitution.

You wanted: \(\Large\rm x=\sec\theta\)\[\Large\rm dx=\sec\theta\tan\theta~d\theta\]

Then,\[\large\rm \int\limits \ln\left(x+\sqrt{x^2-1}\right)dx=\int\limits \ln\left(\sec \theta+\tan \theta\right)\sec \theta \tan \theta~d \theta\]
And you can do parts from here :)
\[\Large\rm u=\ln(\sec \theta+\tan \theta), \qquad \qquad dv=\sec \theta \tan \theta~d \theta\]
The other approach seems good also though. ^^
This way is maybe a bit easier to mess up.

Answer 47

for \[\int\limits \frac{ x }{ \sqrt{x^{2}-1} } dx\]
i got \[\frac{ du }{ 2\sqrt{u} }\]
is that correct?

Answer 48

You should get that BEFORE integrating.
Is that what you meant?

Answer 49

Oh you have a du in there, so yah that would make sense :) lol
Integrate! >.<

Answer 50

is the integration 4u^(1/2)?

Answer 51

ahhh, i made a mistake its u^(1/2) okay i got the answer, thank you all for helping out!!!