Question
Evaluate : $\int_1^2 \frac{\log x}{x^2} \cdot d x$

Answer

$
\begin{aligned}
& \text { Let I }=\int_1^2(\log x)\left(\frac{1}{x^2}\right) \cdot d x \\
&=\left[(\log x) \cdot \int \frac{1}{x^2} \cdot d x\right]_1^2-\int_1^2 \frac{d}{d x} \log x \cdot \int \frac{1}{x^2} \cdot d x \cdot d x \\
&=\left[(\log x) \cdot\left(-\frac{1}{x}\right)\right]_1^2-\int_1^2 \frac{1}{x} \cdot\left(-\frac{1}{x}\right) \cdot d x \\
&=\left[-\frac{1}{x} \log x\right]_1^2+\int_1^2 \frac{1}{x^2} \cdot d x \\
&=\left[-\frac{1}{x} \log x\right]_1^2+\left[-\frac{1}{x}\right]_1^2 \\
&=\left[\left(-\frac{1}{2} \log 2\right)-\left(-\frac{1}{1} \log 1\right)\right]+\left[\left(-\frac{1}{2}\right)-\left(-\frac{1}{1}\right)\right] \\
&=-\frac{1}{2} \log 2-0-\frac{1}{2}+1=\frac{1}{2}-\frac{1}{2} \log 2 \\
& \therefore \quad \int_1^2 \frac{1}{x^2}\left(\frac{\log x}{2}(1-\log 2)\right.
\end{aligned}
$

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