Question

Calcule $\int e^x \cdot \frac{1 - e^{- 2x}}{e^x + e^{- x}} \ dx$.

Answer 1

$\displaystyle\int e^x \cdot \dfrac{1-e^{-2x}}{e^x+e^{-x}}\,dx\\\\\\ =\int \dfrac{e^x\,(1-e^{-2x})}{e^x+e^{-x}}\,dx\\\\\\ =\int \dfrac{e^x-e^x\cdot e^{-2x}}{e^x+e^{-x}}\,dx\\\\\\ =\int \dfrac{e^x-e^{x-2x}}{e^x+e^{-x}}\,dx\\\\\\ =\int \dfrac{e^x-e^{-x}}{e^x+e^{-x}}\,dx~~~~~~\mathbf{(i)}$


Substituição:

$t=e^x+e^{-x}~~\Rightarrow~~\left\{ \begin{array}{l} dt=(e^x-e^{-x})\,dx\\\\ t>0 \end{array} \right.$


Substituindo, a integral $\mathbf{(i)}$ fica

$=\displaystyle\int \dfrac{dt}{t}\\\\\\ =\mathrm{\ell n}|t|+C\\\\ =\mathrm{\ell n}|e^x+e^{-x}|+C\\\\ =\mathrm{\ell n}(e^x+e^{-x})+C$