Question
Evaluate the following integrals:
$\int\text{e}^\text{x}\sqrt{\text{e}^{2\text{x}}+1}\text{dx}$
$\int\text{e}^\text{x}\sqrt{\text{e}^{2\text{x}}+1}\text{dx}$
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Answer
Let $\text{I}=\int\text{e}^\text{x}\sqrt{\text{e}^{2\text{x}}+1}\text{dx}$
Putting $\text{e}^\text{x}=\text{t}$
$\Rightarrow\text{e}^\text{x}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\sqrt{\text{t}^2+1}\text{dt}$
$=\frac{\text{t}}{2}\sqrt{\text{t}^2+1}+\frac{1^2}{2}\ln\Big|\text{t}+\sqrt{\text{t}^2+1}\Big|+\text{C}$
$\Big[\because\ \int\sqrt{\text{x}^2+\text{a}^2}\text{dx}=\frac{1}{2}\text{x}\sqrt{\text{x}^2+\text{a}^2}+\frac{1}{2}\ln\Big|\text{x}+\sqrt{\text{x}^2+\text{a}^2}\Big|+\text{C}\Big]$
$=\frac{\text{e}^\text{x}}{2}\sqrt{\text{e}^{2\text{x}}+1}+\frac{1}{2}\ln\Big|\text{e}^\text{x}+\sqrt{\text{e}^{2\text{x}}+1}\Big|+\text{C}\ \big(\because\ \text{t}=\text{e}^\text{x}\big)$
Putting $\text{e}^\text{x}=\text{t}$
$\Rightarrow\text{e}^\text{x}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\sqrt{\text{t}^2+1}\text{dt}$
$=\frac{\text{t}}{2}\sqrt{\text{t}^2+1}+\frac{1^2}{2}\ln\Big|\text{t}+\sqrt{\text{t}^2+1}\Big|+\text{C}$
$\Big[\because\ \int\sqrt{\text{x}^2+\text{a}^2}\text{dx}=\frac{1}{2}\text{x}\sqrt{\text{x}^2+\text{a}^2}+\frac{1}{2}\ln\Big|\text{x}+\sqrt{\text{x}^2+\text{a}^2}\Big|+\text{C}\Big]$
$=\frac{\text{e}^\text{x}}{2}\sqrt{\text{e}^{2\text{x}}+1}+\frac{1}{2}\ln\Big|\text{e}^\text{x}+\sqrt{\text{e}^{2\text{x}}+1}\Big|+\text{C}\ \big(\because\ \text{t}=\text{e}^\text{x}\big)$
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