Question

integral of dx over sqrtof e^(2x) - 1

Answer 1

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

Answer 2

Looks like you meant to type\[\int \dfrac{dx}{\sqrt{e^{2x}-1}}\]
Let u =e^x
du = e^x dx
1/e^x du = dx
1/u du = dx
Using this substitution, the integral becomes
\[\int \dfrac{\frac{1}{u} du}{\sqrt{u^2-1}}\\
\int \dfrac{du}{u\sqrt{u^2-1}}\]
Look easier?

Answer 3

\[\Large e^x=\sec\alpha \]
Would also maybe help, because

\[\Large \sec^2\alpha -1 = \tan^2 \alpha \]

Answer 4

That substitution definitely works, too. In the end, it's all the same.

Answer 5

\[\Large x= \ln (\sec\alpha) \\
\Large \frac{dx}{d\alpha}=\frac{1}{\sec\alpha}\cdot \tan\alpha \sec\alpha= \tan\alpha \]

Answer 6

Was more a guess, I am actually a bit surprised myself, seems to cancel out very well (-:

Answer 7

sitsh i did actually until there .. and another substitution is needed
trigonometric substitution u=sint for example..
and i get \[\sqrt{-\cos^2 -1}\]

Answer 8

The substitution you should have made was
\[u=\sec t\\
du=\sec t \tan t \;dt\]
The sub you used should be used if you're given an integral containing
\[\sqrt{1-u^2}.\]

Answer 9

i em tryin to find that but at the end i get \[\sqrt{-\cos^2t}\]

Answer 10

and it doestn work so more

Answer 11

thanks .. i did.. with u=sect..