The correct option is D 2 ^ 1 ( e ^ x ) ∫e^xe^2x 1dx Substitute e^x=t ⇒ e^x dx=dt and e^2x=t^2 ∫1t^2 1dt = 2^ 1e^x ...

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Byju's Answer

Standard XII

Mathematics

Parametric Representation: Hyperbola

∫ ex e 2x -1 ...

Question

$\int \frac{e^{x}}{e^{2 x} - 1} d x =$ _____ $+ c$ .

2 {sec}^{- 1} (e^{x})

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{cosec}^{- 1} (e^{x})

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2 {cosec}^{- 1} (e^{x})

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{sec}^{- 1} (e^{x})

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Solution

The correct option is D $2 {sec}^{- 1} (e^{x})$
$\int \frac{e^{x}}{e^{2 x} - 1} d x$

Substitute $e^{x} = t \Rightarrow e^{x} d x = d t$ and $e^{2 x} = t^{2}$

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

$= 2 {sec}^{- 1} e^{x}$

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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

\int \sqrt{\frac{e^{x} - 1}{e^{x} + 1}} d x =

\int \frac{e^{x} - 1}{e^{x} + 1} d x .

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The ans is

= \frac{1}{4} l o g ∣ ∣ ∣ \frac{e^{x} + 1}{e^{x} + k} ∣ ∣ ∣ + C

what is k

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∫exe2x−1dx=_____ +c.

The correct option is D 2sec−1(ex)∫exe2x−1dxSubstitute ex=t⇒exdx=dtand e2x=t2∫1t2−1dt=2sec−1ex

$\int \frac{e^{x}}{e^{2 x} - 1} d x =$ _____ $+ c$ .

The correct option is D $2 {sec}^{- 1} (e^{x})$
$\int \frac{e^{x}}{e^{2 x} - 1} d x$

Substitute $e^{x} = t \Rightarrow e^{x} d x = d t$ and $e^{2 x} = t^{2}$

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

$= 2 {sec}^{- 1} e^{x}$