Question:
Evaluate: $\int\left(e^{x}+\frac{1}{e^{x}}\right)^{2} d x$
Solution:
Let $I=\int\left(e^{x}+\frac{1}{e^{x}}\right)^{2}$
$=\int\left(e^{2 x}+\frac{1}{e^{2 x}}+2\right)$
$=\frac{\mathrm{e}^{2 \mathrm{x}}}{2}-\frac{1}{2} \mathrm{e}^{-2 \mathrm{x}}+2 \mathrm{x}$
Hence, $I=\frac{1}{2} e^{x}+2 x-\frac{1}{2} e^{-2 x}+C$
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