How do you integrate $\frac{ln (x)}{x^{2}}$ $ln(x)/x^2$ ?

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How do you integrate $\frac{ln (x)}{x^{2}}$ $ln(x)/x^2$ ?

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Answer 1 · 2015-06-29T05:51:48

This can be rewritten as:

$\int ln x \cdot \frac{1}{x^{2}} d x$ $int lnx * 1/x^2dx$

Using integration by parts, let:
$u = ln x$ $u = lnx$
$d u = \frac{1}{x} d x$ $du = 1/xdx$
$d v = \frac{1}{x^{2}} d x$ $dv = 1/x^2dx$
$v = - \frac{1}{x} d x$ $v = -1/xdx$

With the formula
$u v - \int v d u$ $uv - intvdu$

$\Rightarrow - \frac{ln x}{x} - \int - \frac{1}{x} \cdot \frac{1}{x} d x$ $=> -lnx/x - int -1/x*1/xdx$

$= - \frac{ln x}{x} - \int - \frac{1}{x^{2}} d x$ $= -lnx/x - int -1/x^2dx$

$= - \frac{ln x}{x} - \frac{1}{x} + C$ $= color(blue)(-lnx/x - 1/x + C)$

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How do you integrate $\frac{ln (x)}{x^{2}}$ $ln(x)/x^2$ ?

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