Assume that \(f(X)\) is a rational function of X. Then, in the case of \(f(\mathrm{e}^{x})\), the substitution \(\mathrm{e}^{x}=t\) will produce an integral of a rational faction of t. That is: $$\int f(\mathrm{e}^{x})\ dx=\int f(t) \frac{1}{t}\ dt$$
Example : Evaluate \(\int \frac{\mathrm{e}^{x}-1}{\mathrm{e}^{2x}-1}\ dx\)
Solution :
If we let \(\mathrm{e}^{x}=t\), we have \(dx=\frac{1}{\mathrm{e}^{x}}\ dt=\frac{1}{t}\ dt\).
Thus, the Substitution Rule gives
$$\begin{eqnarray*}&&\ \int \frac{\mathrm{e}^{x}-1}{\mathrm{e}^{2x}-1}\ dx=\int \frac{t-1}{t^{2}+1}\cdot \frac{1}{t}\ dt =\int \left(\frac{t+1}{t^{2}+1}-\frac{1}{t}\right)=\frac{1}{2}\ln (t^{2}+1)+\tan^{-1}t-\ln t+C\end{eqnarray*}$$
Note: To express the rational function \(\frac{t-1}{(t^{2}+1)t}\) as a sum of partial fractions, we represent in the following form:
$$\frac{t-1}{(t^{2}+1)t}=\frac{at+b}{t^{2}+1}+\frac{c}{t}$$
Multiplying by the least common denominator \((t^{2}+1)t\), and equating coefficients, we obtain \(a=1, b=1\) and \(c=-1\). See also integrals of rational functions.