Question
Evaluate the following integrals:
$\int\frac{\text{e}^{2\text{x}}}{1+\text{e}^\text{x}}\text{ dx}$

Answer

$\int\frac{\text{e}^{2\text{x}}}{1+\text{e}^\text{x}}\text{ dx}$

$\Rightarrow\int\frac{\text{e}^\text{x}.\text{e}^\text{x}}{1+\text{e}^\text{x}}\text{ dx}$ then,

Let $1+\text{e}^\text{x}=\text{t}$

$\Rightarrow\text{e}^\text{x}=\text{t}-1$

$\Rightarrow\text{e}^\text{x}\text{dx}=\text{dt}$

Now, $\int\frac{\text{e}^\text{x}.\text{e}^\text{x}}{1+\text{e}^\text{x}}\text{ dx}$

$=\int\frac{(\text{t}-1)\text{dt}}{\text{t}}$

$=\Big(1-\frac{1}{\text{t}}\Big)\text{dt}$

$=\text{t}-\log|\text{t}|+\text{C}$

$=\big(1+\text{e}^\text{x}\big)-\log\big(1+\text{e}^\text{x}\big)+\text{C}$

Let $\text{C}+1=\text{C}^\text{n}$

$=\text{e}^\text{x}-\log\big(1+\text{e}^\text{x}\big)+\text{C}^\text{n}$

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