I=∫log xxdx Let log x=t Differentiating wrt x 1xdx=dt Therefore, I=∫ t dt By using ∫x^ndx=x^n+1n+1+c I=t^22+C I=(log x)^ ...

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Special Integrals - 1

Evaluate: ∫l...

Question

Evaluate:
$\int \frac{log x}{x} d x$

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Solution

$I = \int \frac{log x}{x} d x$

Let $log x = t$

Differentiating wrt $x$

$\frac{1}{x} d x = d t$

Therefore,

$I = \int t d t$

By using $\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c$

$I = \frac{t^{2}}{2} + C$

$I = \frac{(log x)^{2}}{2} + C$

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