Find: e^ x (2 + e^ x ) (4 + e^ 2x ) dx. - Vidyadip

Question

Find:
$\int \frac{\text{e}^{\text{x}}}{(2 + \text{e}^{\text{x}}) (4 + \text{e}^{2\text{x}})} \text{dx}.$

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Answer

$\text{I} = \int \frac{\text{dt}}{\text{(2 + t) (4 +} \text{t}^{2})} \text{ } \text{where} \text{ e}^{\text{x}} = \text{t}$
$\text{Now}, \frac{1}{\text{(2 + t) (4 + t}^{2})} = \frac{1}{8 (2 + \text{t})} - \frac{1}{8} \bigg(\frac{\text{t - 2}}{\text{4 + t}^{2}}\bigg)$
$\Rightarrow \int \frac{\text{dt}}{\text{(2 + t) (4 + t}^{2})} = \frac{1}{8} \log | \text{2 + t|} = \frac{1}{16} \log |\text{4 + t}^{2}| + \frac{1}{8} \tan^{-1} \bigg(\frac{\text{t}}{2}\bigg) + \text{c}$
$\Rightarrow \int \frac{\text{e}^{\text{x}}\text{dx}}{(2 + \text{e}^{\text{x}}) (4 + \text{e}^{2\text{x}})} = \frac{1}{8} \log |\text{2 + e}^{\text{x}}| - \frac{1}{16} \log \text{|4 + e}^{\text{2x}}| + \frac{1}{8} \tan^{-1} \bigg(\frac{\text{e}^{\text{x}}}{2}\bigg) + \text{c}$

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