Question
Evaluate the following integrals:
$\int\frac{\sqrt{16+(\log\text{x})^2}}{\text{x}}\text{dx}$
$\int\frac{\sqrt{16+(\log\text{x})^2}}{\text{x}}\text{dx}$
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Answer
Let $\text{I}=\int\frac{\sqrt{16+(\log\text{x})^2}}{\text{x}}\text{dx}$
Let $\log\text{x}=\text{t}$
$\Rightarrow\frac{1}{\text{x}}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\sqrt{16+\text{t}^2}\text{dt}$
$=\int\sqrt{4^2+\text{t}^2}\text{dt}$
$=\frac{\text{t}}{2}\sqrt{16+\text{t}^2}+\frac{16}{2}\log\big|\text{t}+\sqrt{16+\text{t}^2}\big|+\text{C}$
$\therefore\ \text{I}=\frac{\log\text{x}}{2}\sqrt{16+(\log\text{x})^2}\\+8\log\Big|\log\text{x}+\sqrt{16+(\log\text{x})^2}\Big|+\text{C}$
Let $\log\text{x}=\text{t}$
$\Rightarrow\frac{1}{\text{x}}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\sqrt{16+\text{t}^2}\text{dt}$
$=\int\sqrt{4^2+\text{t}^2}\text{dt}$
$=\frac{\text{t}}{2}\sqrt{16+\text{t}^2}+\frac{16}{2}\log\big|\text{t}+\sqrt{16+\text{t}^2}\big|+\text{C}$
$\therefore\ \text{I}=\frac{\log\text{x}}{2}\sqrt{16+(\log\text{x})^2}\\+8\log\Big|\log\text{x}+\sqrt{16+(\log\text{x})^2}\Big|+\text{C}$
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