Question

( e ^ { x } + e ^ { - x } ) ^ { \frac { 1 } { 4 } } ( e ^ { x } - e ^ { - x } ) d x

Answer 1

$\displaystyle\textsf{Let}\hspace{0.1cm}I = \int (e^{x} + e^{-x})^{\frac{1}{4}} (e^{x} - e^{-x}) \hspace{0.1cm}\mathrm{d}x\\ \displaystyle\textsf{Let}\hspace{0.1cm} u = e^{x} + e^{-x}\\ \frac{\mathrm{d}u}{\mathrm{d}x} = e^{x} - e^{-x}\\ \displaystyle\frac{\mathrm{d}u}{e^{x} - e^{-x}} = \mathrm{d}x\\ \displaystyle I = \int u^{\frac{1}{4}} (\cancel{e^{x} - e^{-x}}) \frac{\mathrm{d}u}{\cancel{e^{x} - e^{-x}}}\\ \displaystyle\begin{aligned} I &= \int u^{\frac{1}{4}}\hspace{0.1cm}\mathrm{d}u \\&= \frac{4u^{\frac{5}{4}}}{5} + C \end{aligned}\\ \therefore I = \frac{4(e^{x} + e^{-x})^{\frac{5}{4}}}{5} + C$