How do you calculate $int\ ln(x)^ln(x)(1/x+ln(ln(x))/x)\ dx$ ?

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Answer 1 · 2018-01-23T12:25:39

$int\ ln(x)^ln(x)(1/x+ln(ln(x))/x)\ dx=ln(x)^ln(x)+C$

$\ \ \ \ \ \ int\ ln(x)^ln(x)(1/x+ln(ln(x))/x)\ dx$

Let's first substitute $u=ln(x)$ and $du=dx/x$ :
$=int\ u^u(1+ln(u))\ du$
$=int\ e^(u ln(u))(1+ln(u))\ du$

Substitute again with $v=u ln(u)$ and $dv=(1+ln(u))\ du$ :
$=int\ e^v\ dv$
$=e^v+C$

Substitute back $v=u ln(u)$ :
$=e^(u ln(u))+C$
$=u^u+C$

Substitute back $u=ln(x)$ :
$=ln(x)^ln(x)+C$

How do you calculate int\ ln(x)^ln(x)(1/x+ln(ln(x))/x)\ dxint\ ln(x)^ln(x)(1/x+ln(ln(x))/x)\ dx ?