The correct option is C None of these ∫e^2x 1e^2x + 1dx = ∫ (e^x e^ xe^x + e^ x )dx = log (e^x + e^ x) + c = log (e^x + 1) log e^x + c = x ...

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Theorems on Integration

∫e2x - 1e2x +...

Question

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

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

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

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log \frac{e^{2 x} + e^{- 2 x}}{e^{2 x} - e^{- 2 x}} + c

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None of these

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Solution

The correct option is C None of these
$\int \frac{e^{2 x} - 1}{e^{2 x} + 1} d x = \int (\frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}) d x$
$= log (e^{x} + e^{- x}) + c = log (e^{x} + 1) - log e^{x} + c$
$= - x + log (e^{2 x} + 1) + c$
$∴$ Choice (d) is correct.

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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. Differentiate

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

w.r.t

x

Q. Find the differential coefficient with respect to

x

\frac{e^{2 x} + e^{- 2 x}}{e^{2 x} - e^{- 2 x}}

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)

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∫e2x−1e2x+1dx is

The correct option is C None of these∫e2x−1e2x+1dx=∫(ex−e−xex+e−x)dx=log(ex+e−x)+c=log(ex+1)−logex+c=−x+log(e2x+1)+c∴ Choice (d) is correct.

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

The correct option is C None of these
$\int \frac{e^{2 x} - 1}{e^{2 x} + 1} d x = \int (\frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}) d x$
$= log (e^{x} + e^{- x}) + c = log (e^{x} + 1) - log e^{x} + c$
$= - x + log (e^{2 x} + 1) + c$
$∴$ Choice (d) is correct.