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

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

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log \frac{e^{x} + e^{- x}}{e^{x} - e^{- x}} + C

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

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

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Solution

The correct option is C $log | e^{x} + e^{- x} | + C$
Let $I = \int \frac{e^{2 x} - 1}{e^{2 x} + 1} d x$

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

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

$= \int \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} \cdot d x$
Let $(e^{x} + e^{- x}) = t$ $\Rightarrow (e^{x} - e^{- x}) d x = d t$
$\Rightarrow d x = \frac{d t}{e^{x} - e^{- x}}$

$∴ I = \int \frac{e^{x} - e^{- x}}{t} \cdot \frac{d t}{e^{x} - e^{- x}}$
$= \int \frac{d t}{t}$
$= log | t | + C$
$= log | e^{x} + e^{- x} | + C$

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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

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

(a)

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

(b)

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(d)

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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)

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

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

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∫e2x−1e2x+1dx is equal to

The correct option is C log|ex+e−x|+CLet I=∫e2x−1e2x+1dx=∫(ex⋅ex−1)(ex⋅ex+1)dx=∫ex(ex−1ex)ex(ex+1ex)⋅dx=∫ex−e−xex+e−x⋅dxLet (ex+e−x)=t ⇒(ex−e−x)dx=dt⇒dx=dtex−e−x∴I=∫ex−e−xt⋅dtex−e−x=∫dtt=log|t|+C =log|ex+e−x|+C

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