Question

what is the integration of 1/ sqrt (1- e^2x) ? The answer given to this question is in terms of log and i am unable to do it. please help at the earliest. Thanks a tonne.

Answer 1

Hope this is the solution you are looking for. Have a nice day:)

Answer 2

Answer:

$\frac{1}{2}\times log[\frac{\sqrt(1 -1 - e^2x)}{\sqrt(1 +1 - e^2x)}] + C$

Step-by-step explanation:

According to this question

We have been given that

$\int \frac{1}{\sqrt(1 - e^2x)}dx$

Let,  $1 - e^2x$ = $t^2$

$- e^2x$ = $t^2 - 1$

Differentiation both side

$-2e^2x$ = $2t\frac{dt}{dx}$

$dx$ = $\frac{t}{-e^2x}\times dt$

Then,

$\int \frac{1}{\sqrt {t^2}}\times \frac{t}{t^2 - 1}dt$

$\int \frac{1}{t}\times \frac{t}{t^2 - 1}$

$\int \frac{1}{t^2 - 1}$

$\frac{1}{2}\times log[\frac{1 - t}{1 + t}] + C$

$\frac{1}{2}\times log[\frac{\sqrt(1 -1 - e^2x)}{\sqrt(1 +1 - e^2x)}] + C$