$ \int\frac{g(x)}{f(x)} \, dx $ where $f(x) = \frac{1}{2}(e^x+e^{-x})$ and $g(x) = \frac{1}{2}(e^x - e^{-x})$?

integration indefinite-integrals

Solution 1:

Note that if $K$ is a constant, $\ln|K\cdot h(x)|= \ln|K|+\ln|h(x)|$. In other words, the $1/2$ in the log can get 'absorbed' into the $+C$ term.

Solution 2:

Using the identity: $ \log (\frac{a}{b})=\log(a)- \log(b)$

$\ln \frac{1}{2}(e^x+e^{-x})+C$
$=\ln (e^x+e^{-x})-\ln 2+C $
$=\ln (e^x+e^{-x})+C{'} $ (as $\ln 2$ is constant)

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