The correct option is A log (e^x+e^ x)+C ∫( e^2x 1)e^2x + 1dx ∫e^x( e^x e^ x)e^x( e^x + e^ x) dx( takee^x common from both Numerator and ...

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Special Integrals - 1

∫ e2x - 1/e2x...

Question

$\int \frac{(e^{2 x} - 1)}{e^{2 x} + 1} d x$

log (e^{x} + e^{- x}) + C

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log (e^{x} - e^{- x}) + C

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log (e^{2 x} + e^{- 2 x}) + C

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log (e^{2 x} - e^{- 2 x}) + C

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Solution

The correct option is A $log (e^{x} + e^{- x}) + C$
$\int \frac{(e^{2 x} - 1)}{e^{2 x} + 1} d x$

$\int \frac{e^{x} (e^{x} - e^{- x})}{e^{x} (e^{x} + e^{- x})} d x (t a k e e^{x} common from both Numerator and Denaminator) .$

$\int \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} d x$

$l e t e^{x} + e^{- x} = t$

$(e^{x} - e^{- x}) d x = d t$

$= \int \frac{d t}{t}$

$= log t + c$

$= log (e^{x} + e^{- x}) + c$

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Similar questions

Let I $= \int \frac{e^{x}}{e^{4 x} + e^{2 x} + 1} d x . J = \int \frac{e^{- x}}{e^{- 4 x} + e^{- 2 x} + 1} d x,$ Then, for an arbitrary constant c, the value of J-I equals

Q. Let

I = \int \frac{e^{x}}{e^{4 x} + e^{2 x} + 1} d x, J = \int \frac{e^{- x}}{e^{- 4 x} + e^{- 2 x} + 1} d x .

Then, for an arbitrary constant

C

, the value for

J - I

equals

Q. Let

I = \int \frac{e^{x}}{e^{4 x} + e^{2 x} + 1} d x, J = \int \frac{e^{- x}}{e^{- 4 x} + e^{- 2 x} + 1} d x .

Then the value for

J - I

equals to
( where

C

is integration constant)

Let I $= \int \frac{e^{x}}{e^{4 x} + e^{2 x} + 1} d x . J = \int \frac{e^{- x}}{e^{- 4 x} + e^{- 2 x} + 1} d x,$ Then, for an arbitrary constant c, the value of J-I equals

Q. Differentiate

y = [\frac{e^{2 x} + e^{- 2 x}}{e^{2 x} - e^{- 2 x}}]

w.r.t

x

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∫(e2x−1)e2x+1dx

The correct option is A log(ex+e−x)+C∫(e2x−1)e2x+1dx∫ex(ex−e−x)ex(ex+e−x)dx(take ex common from both Numerator and Denaminator).∫ex−e−xex+e−xdxlet ex+e−x=t(ex−e−x)dx=dt=∫dtt=logt + c=log(ex+e−x)+c

$\int \frac{(e^{2 x} - 1)}{e^{2 x} + 1} d x$