Question

Dum hai to solve kar ke dikhaoo...
Integrate: 1/square root e^2x-1

Answer 1

Step-by-step explanation:

Integral ( 1/ sqrt (e^(2x) - 1) dx )

Let u = sqrt ( e^(2x) - 1 ). Then

u^2 = e^(2x) - 1

Our goal here is to isolate the x, so that we may obtain a calculation for dx.

u^2 + 1 = e^(2x)

Taking the natural log of both sides, we get

ln (u^2 + 1) = 2x

Multiplying both sides by (1/2),

(1/2) ln (u^2 + 1) = x

Differentiating,

(1/2) (1/(u^2 + 1))(2u) du = dx

This makes our integral

Integral ( (1/u) (1/2) (1/(u^2 + 1))(2u) du )

Cancelling terms (i.e 1/u will cancel with the u next to the du), and pulling out constants, we get

(1/2)(2) * Integral ( 1/(u^2 + 1) du )

Integral ( 1/(u^2 + 1) du )

This is a known derivative; it's the derivative of arctan. Therefore, we get

arctan(u) + C

But u = sqrt(e^(2x) - 1), so our final answer is

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

Le kar diya.... abb moj mana...