Integrate the rational function in exercise: 1 (e^x-1) [Hint: Put… - Vidyadip

Question

Integrate the rational function in exercise:
$\frac{1}{(\text{e}^\text{x}-1)}$
$[$Hint: Put $e^x = t]$

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Answer

$\text{I}=\int\frac{1}{\text{e}^\text{x}-1}\text{dx}\dots(\text{i})$ Putting $e^x = t$
$\Rightarrow \ \text{e}^\text{x}=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow \text{e}^{\text{x}} \text{dx = dt} \Rightarrow \text{x}$$\Rightarrow \ \text{dx}=\frac{\text{dt}}{\text{e}^\text{x}}$
$\therefore$ From eq. (i), $\text{I}=\int\frac{1}{\text{t}-1}\frac{\text{dt}}{\text{e}^\text{x}}=\int\frac{1}{\text{t}-1}\frac{\text{dt}}{\text{t}}=\int\frac{1}{\text{t}(\text{t}-1)}\text{dt}=\int\frac{\text{t}-(\text{t}-1)}{\text{t}(\text{t}-1)}\text{dt}$ $=\int\Bigg(\frac{\text{t}}{\text{t}(\text{t}-1)}-\frac{(\text{t}-1)}{\text{t}(\text{t}-1)}\Bigg)\text{dt}=\int\Bigg(\frac{1}{(\text{t}-1)}-\frac{1}{\text{t}}\Bigg)\text{dt}=\int\frac{1}{\text{t}-1}\text{dt}-\int\frac{1}{\text{t}}\text{dt}$$=\text{log}|\text{t}-1|-\text{log}|\text{t}|+\text{c}=\text{log}\Bigg|\frac{\text{t}-1}{\text{t}}\Bigg|+\text{c}=\text{log}\Bigg|\frac{\text{e}^\text{x}-1}{\text{e}^\text{x}}\Bigg|+\text{c}$

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