Question

help please

Answer 1

\[\int\limits_{0}^{\ln10}\frac{ (e^x)(\sqrt{e^x-1}) }{ (e^x+8) }\]

Answer 2

You use too many unnecessary parentheses\[\large \int\limits\limits_{0}^{\ln10}\frac{ e^x\sqrt{e^x-1} }{ e^x+8 } dx\]

Answer 3

ok lool

Answer 4

so then

Answer 5

Then idk...
http://www.wolframalpha.com/input/?i=%5Cint+%5Cfrac%7B+e%5Ex%5Csqrt%7Be%5Ex-1%7D+%7D%7B+e%5Ex%2B8+%7D+dx

Answer 6

wym idk lmao

Answer 7

lol i mean idk.\[\large \int\limits \frac{ e^x\sqrt{e^x-1} }{ e^x+8 } dx\]try some shiz like \(u=e^x - 1\) so \(du = e^x dx\)\[\large \int\limits \frac{ \sqrt{u} }{ u+9 } du\]

Answer 8

iight I think i got it now, with looking at wolframalpha for help but it ain't finna be easy\[\large \int\limits\limits \frac{ \sqrt{u} }{ u+9 } du \]Let \(\large w = \sqrt u \)\[\large dw = \frac{ 1 }{ 2 \sqrt u } du\]\[\large 2 \sqrt u \text{dw} = du\]or\[\large 2 w \text{dw} = du\]\[\large 2\int\limits \frac{ w^2 }{ w^2+9 } dw\]Now it's actually solvable but you still gotta do long division...

Answer 9

Or you can use (11) http://integral-table.com/downloads/single-page-integral-table.pdf

Answer 10

woahhh lol

Answer 11

Yeah it's an ugly azz integral... https://www.youtube.com/watch?v=b1wzQNzttSk

Answer 12

yaas lool hmm

Answer 13

help me with few more lmao x'd

Answer 14

\[\large 2\int\limits\limits \frac{ w^2 }{ w^2+9 } dw = 2(w-3\tan^{-1}\frac{ w }{ 3 })\]then unsubstitute\[\large 2(\sqrt u -3\tan^{-1}\frac{ \sqrt u }{ 3 })\]
\[\large 2(\sqrt {e^x - 1} -3\tan^{-1}\frac{ \sqrt {e^x - 1} }{ 3 })\]

Answer 15

Then you gotta plug shet in, them limits.

Answer 16

iight

Answer 17

iight ill do that when im done x'd wanan do my problems neega lmaoo

Answer 18

lol just plug in WA if you want em done... http://www.wolframalpha.com/input/?i=%5Cint%5Climits_%7B0%7D%5E%7B%5Cln+10%7D%5Cfrac%7B+e%5Ex%5Csqrt%7Be%5Ex-1%7D+%7D%7B+e%5Ex%2B8+%7D+dx

Answer 19

lool truee

Answer 20

is there an easier way to do comparison theorem problems

Answer 21

Depends what you think is a hard way lol. Post one

Answer 22

lmao idk leh me post it
number 4 a and b

Answer 23

For a, sin x is always between 1 and -1, so \[\large \frac{ 2+ \sin x }{\sqrt x } \ge \frac{ 1 }{ \sqrt x }\]And 1/sqrt x will be divergent (which means anything bigger diverges) because an integral of \(\Large \frac{ 1 }{ x^c }\) only converges if the exponent is greater than 1.

Answer 24

For b, because \[\large \sqrt{1+x^4} \ge \sqrt{x^4}\]then\[\large \int\limits_{1}^{\infty} \frac{ 1 }{ \sqrt{1+x^4}}dx \le \int\limits_{1}^{\infty} \frac{ 1 }{ x^2}dx\](bigger denom means smaller fraction)

and that right one converges for the reason i gave above, which means anything smaller converges.

Answer 25

so how u knows its gonna be a smaller fraction ?

Answer 26

cos... all the things i just said lol

Answer 27

oh wait yeh x'd lmao

Answer 28

\[\large 1+x^4\ge x^4\] so when you square root both sides, that's still true

Answer 29

dangg u smart ;p

Answer 30

loool ty neega

Answer 31

x'd

Answer 32

x'd

Answer 33

x'd iight number 5

Answer 34

Shouldn't be that hard hopefully. Just use the arc length formula

Answer 35

so the derivative of that is cot (1/2x) ?

Answer 36

Yeh

Answer 37

Now do the arc formula and you can use the 1+cot^2 identity