Evaluate: 1 x -1 ( x)^2 d x; (where .x>1 ). - Vidyadip

Question

Evaluate: $\int\left\{\frac{1}{\log x}-\frac{1}{(\log x)^2}\right\} d x; ($where $\left.x>1\right)$.

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Answer

$ \int\left\{\frac{1}{\log _e x}-\frac{1}{\left(\log _e x\right)^2}\right\} d x$
$=\int \frac{d x}{\log _e x}-\int \frac{1}{\left(\log _e x\right)^2} d x=\frac{1}{\log _e x} \int d x-\int\left\{\frac{d}{d x}\left(\frac{1}{\log _e x}\right) \int d x\right\} d x-\int \frac{1}{\left(\log _e x\right)^2} d x$
$=\frac{x}{\log _e x}+\int \frac{1}{\left(\log _e x\right)^2} \frac{1}{x} \cdot x \cdot d x-\int \frac{1}{\left(\log _e x\right)^2} d x$
$=\frac{x}{\log _e x}+\int \frac{1}{\left(\log _e x\right)^2} d x-\int \frac{d x}{\left(\log _e x\right)^2}=\frac{x}{\log _e x}+c $where' $c$ 'is any arbitary constant of integration.

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