Question

evaluate the integral

Answer 1

This is is a bit tricky for me

Answer 2

\[\int\limits_{}^{}\sqrt{e ^{2x}-1}dx\]

Answer 3

so I know you set u = to \[\sqrt{e ^{2x}-1}\]

Answer 4

for some reason a I am blanking out how to find du

Answer 5

I am guessing u substitution to find du

Answer 6

Looks good! keep going :)

Answer 7

\(\large u = \sqrt{e ^{2x}-1}\)

\(\large du = \dfrac{1}{2\sqrt{e ^{2x}-1}}\cdot e^{2x}\cdot 2 ~dx\)

Answer 8

can you show me how you got that

Answer 9

oh chain rule

Answer 10

whats the derivative of \(\sqrt{x}\) with respect to \(x\) ?

Answer 11

yes :)

Answer 12

\(\large u = \sqrt{e ^{2x}-1}\)

\(\large du = \dfrac{1}{2\sqrt{e ^{2x}-1}}\cdot e^{2x}\cdot 2 ~dx = \dfrac{e^{2x}}{\sqrt{e ^{2x}-1}}dx\)

Answer 13

\(\large u = \sqrt{e ^{2x}-1}\)

\(\large du = \dfrac{1}{2\sqrt{e ^{2x}-1}}\cdot e^{2x}\cdot 2 ~dx = \dfrac{e^{2x}}{\sqrt{e ^{2x}-1}}dx = \dfrac{u^2+1}{u}d\)

\(\large \implies \dfrac{udu}{u^2+1} = dx \)

Answer 14

okay, let me right this down.

Answer 15

u sub again?

Answer 16

the integral becomes
\[\large \int\sqrt{e ^{2x}-1}dx=\int u \dfrac{udu}{u^2+1} = \int \dfrac{u^2}{u^2+1}du\]

Answer 17

which should be easy to integrate

Answer 18

okay thanks I can take it from here =)

Answer 19

http://www.wolframalpha.com/input/?i=%5Cint+%5Csqrt%7Be%5E%7B2x%7D-1%7Ddx

Answer 20

you may use wolfram to double check ur answer ^

Answer 21

sqrt(e^2x-1) + arctan(sqrt(e^2x - 1)) + C

Answer 22

sorry a bit messy

Answer 23

oh just saw the wolfram

Answer 24

wolfram says

sqrt(e^2x-1) \(\color{red}{-}\) arctan(sqrt(e^2x - 1)) + C

Answer 25

its - arctan?

Answer 26

yes