I=∫e^x e^ x/e^x+e^ xdx Let (e^x+e^ x)=f (e^x e^ x)dx=df (e^x e^ x)dx=df ∴ I =∫df/f = log f+C substituting value of 1 ∴ I=lo ...

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Integration by Substitution

∫ex-e-xex+e-x

Question

$\int \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}$

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Solution

$I = \int \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} d x$
Let - $(e^{x} + e^{- x}) = f$
$(e^{x} - e^{- x}) d x = d f$
$(e^{x} - e^{- x}) d x = d f$
$∴ I = \int \frac{d f}{f}$
$= log f + C$
substituting value of 1-
$∴ I = log | e^{x} + e^{- x} | + C$

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