- dx - e 2x -1 equals to

The correct option is A ln|e^x 1| x+C Given, ∫1√(e^2x) 1dx =∫1e^x 1dx =∫e^xe^x 1 1dx as 1e^x 1=e^xe^x 1 1 =∫e^xe^x 1dx ∫ 1dx #16 ...

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Byju's Answer

Standard XII

Mathematics

Rationalization Method to Remove Indeterminate Form

∫ dx √ e 2x -...

Question

$\int \frac{d x}{\sqrt{e^{2 x}} - 1}$ equals to

ln | e^{x} - 1 | - x + C

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{cos}^{- 1} e^{x}

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ln | e^{x} | - x + C

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

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Solution

The correct option is A $ln | e^{x} - 1 | - x + C$
Given,

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

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

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

as $\frac{1}{e^{x} - 1} = \frac{e^{x}}{e^{x} - 1} - 1$

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

$= ln | e^{x} - 1 | - x + C$

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Similar questions

Q. If

l n = \int (1 + x + x^{- 1}) e^{x + x^{- 1}} d x =

Q. Let

\frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} = l n \sqrt{\frac{1 + x}{1 - x}},

then find

x

\int e^{x} {(\frac{1 - x}{1 + x^{2}})}^{2} d x is equal to (a) \frac{e^{x}}{1 + x} + C (b) - \frac{e^{x}}{1 + x^{2}} + C (c) \frac{e^{x}}{(1 + x}

²)2+C (d) -ex(1+x²)2+C

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

is equal to

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∫dx√e2x−1 equals to

The correct option is A ln|ex−1|−x+CGiven,∫1√e2x−1dx=∫1ex−1dx=∫exex−1−1dxas 1ex−1=exex−1−1=∫exex−1dx−∫1dx =ln|ex−1|−x+C

$\int \frac{d x}{\sqrt{e^{2 x}} - 1}$ equals to