Find the following integral. $$ \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} d x $$ $$ \int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} d x=\square $$

Question

UpStudy AI · Accepted Answer

We start with the integral

$$
\int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\,dx.
$$

Notice that the numerator and denominator can be written in terms of hyperbolic functions:

$$
\begin{aligned}
\sinh x &= \frac{e^x - e^{-x}}{2}, \
\cosh x &= \frac{e^x + e^{-x}}{2}.
\end{aligned}
$$

Thus, the integrand becomes

$$
\frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{2\sinh x}{2\cosh x} = \tanh x.
$$

So the integral simplifies to

$$
\int \tanh x\,dx.
$$

The antiderivative of $$\tanh x$$ is known. One way to verify is by recalling that

$$
\frac{d}{dx}\ln(\cosh x)=\frac{1}{\cosh x}\sinh x=\tanh x.
$$

Thus, we have

$$
\int \tanh x\,dx = \ln(\cosh x) + C.
$$

Finally, the evaluated integral is

$$
\int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\,dx = \ln(\cosh x) + C.
$$
The integral evaluates to $$ \ln(\cosh x) + C $$.

Find the following integral. \[ \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} d x \] \( \int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} d x=\square \)