Question
$\text{Find:}\int \frac{\text{x}\ \text{dx}}{(2 + \text{x}^{\text{2}}) (4 + \text{e}^{4\text{}})}$
✓
Answer
$\text{I} =\frac{1}{2} \int \frac{\text{1}}{\text{(2 + t) (4 +} \text{t}^{2})}\text{dt} \text{ } \text{where} \ \text{t} = \text{x}^2$
$\text{Now},\Bigg[ \frac{1}{\text{(2 + t) (4 + t}^{2})}\Bigg] =\frac{1}{2}\Bigg[ \frac{1}{8 (2 + \text{t})} - \frac{1}{8} \bigg(\frac{\text{t - 2}}{\text{4 + t}^{2}}\bigg)\Bigg]$
$\Rightarrow \int \frac{\text{1}}{\text{(2 + t) (4 + t}^{2})} \text{dt}= \frac{1}{16} \log | \text{2 + t|} = -\frac{1}{32} \log |\text{4 + t}^{2}| + \frac{1}{16} \tan^{-1} \bigg(\frac{\text{t}}{2}\bigg) + \text{c}$
$\Rightarrow \int \frac{\text{x}}{(2 + \text{x}^{\text{2}}) (4 + \text{x}^{4\text{}})}\text{dx} = \frac{1}{16} \log |\text{2 + x}^{\text{2}}| - \frac{1}{32} \log \text{|4 + x}^{\text{4}}| + \frac{1}{16} \tan^{-1} \bigg(\frac{\text{x}^{\text{2}}}{2}\bigg) + \text{c}$
$\text{Now},\Bigg[ \frac{1}{\text{(2 + t) (4 + t}^{2})}\Bigg] =\frac{1}{2}\Bigg[ \frac{1}{8 (2 + \text{t})} - \frac{1}{8} \bigg(\frac{\text{t - 2}}{\text{4 + t}^{2}}\bigg)\Bigg]$
$\Rightarrow \int \frac{\text{1}}{\text{(2 + t) (4 + t}^{2})} \text{dt}= \frac{1}{16} \log | \text{2 + t|} = -\frac{1}{32} \log |\text{4 + t}^{2}| + \frac{1}{16} \tan^{-1} \bigg(\frac{\text{t}}{2}\bigg) + \text{c}$
$\Rightarrow \int \frac{\text{x}}{(2 + \text{x}^{\text{2}}) (4 + \text{x}^{4\text{}})}\text{dx} = \frac{1}{16} \log |\text{2 + x}^{\text{2}}| - \frac{1}{32} \log \text{|4 + x}^{\text{4}}| + \frac{1}{16} \tan^{-1} \bigg(\frac{\text{x}^{\text{2}}}{2}\bigg) + \text{c}$
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