Integrate the following functions- e 2 x-1- e 2 x-1 dx

∫e^2x 1/e^2x+1dx=∫(e^x.e^x 1)/(e^x.e^x +1)dx =∫e^x (e^x 1/e^x)/e^x (e^x +1/e^x)dx =∫(e^x e^ x)/e^x +e^ xdx Let e^x +e^ x=t ⇒ e^x e^ x=dt/dx⇒ dx =dt/e^x e^ x ...

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Byju's Answer

Standard XII

Mathematics

Integration by Partial Fractions

Integrate the...

Question

Integrate the following functions.
$\int \frac{e^{2 x} - 1}{e^{2 x} + 1} d x .$

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Solution

$\int \frac{e^{2 x} - 1}{e^{2 x} + 1} d x = \int \frac{(e^{x} . e^{x} - 1)}{(e^{x} . e^{x} + 1)} d x = \int \frac{e^{x} (e^{x} - \frac{1}{e^{x}})}{e^{x} (e^{x} + \frac{1}{e^{x}})} d x = \int \frac{(e^{x} - e^{- x})}{e^{x} + e^{- x}} d x$
Let $e^{x} + e^{- x} = t \Rightarrow e^{x} - e^{- x} = \frac{d t}{d x} \Rightarrow d x = \frac{d t}{e^{x} - e^{- x}}$
$\int \frac{e^{2 x} - 1}{e^{2 x} + 1} d x = \int \frac{(e^{x} - e^{- x})}{t} \frac{d t}{e^{x} - e^{- x}} = \int \frac{1}{t} d t = l o g | t | + C = l o g | e^{x} + e^{- x} | + C$

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Similar questions

Integrate the following functions.
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\int \frac{e^{2 x} - 1}{e^{2 x} + 1} d x

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Integrate the following functions. ∫e2x−1e2x+1dx.

∫e2x−1e2x+1dx=∫(ex.ex−1)(ex.ex+1)dx=∫ex(ex−1ex)ex(ex+1ex)dx=∫(ex−e−x)ex+e−xdx Let ex+e−x=t⇒ex−e−x=dtdx⇒dx=dtex−e−x ∫e2x−1e2x+1dx=∫(ex−e−x)tdtex−e−x=∫1tdt=log|t|+C=log|ex+e−x|+C

Integrate the following functions.
$\int \frac{e^{2 x} - 1}{e^{2 x} + 1} d x .$