How do you evaluate the integral of $int ln(sqrt x)/x$ ?

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Answer 1 · 2016-02-15T00:44:23

Use the fact that $ln(sqrtx) = ln(x^(1/2))=1/2lnx$

So we have:

$int ln(sqrt x)/x = 1/2 int lnx 1/xdx$

Because $d/dx(lnx) = 1/x$ , we have

$1/2 int u du = 1/4u^2 +C$

$= 1/4(lnx)^2 +C$

Check the answer by differentiating.

How do you evaluate the integral of int ln(sqrt x)/xint ln(sqrt x)/x?