The value of the definite integral ∫ limits e e 2010 (1/ x )(1+(1- ln x / ln x ln (( x ) ln x ))) dx, equals

Question

The value of the definite integral ∫ limits e e 2010 (1/ x )(1+(1- ln x / ln x ln (( x ) ln x ))) dx, equals (A) 2009- ln (2010- ln 2010) (B) 2010- ln (2009- ln 2009) (C) 2009- ln (2010- ln 2009) (D) 2010- ln (2010- ln 2010)

Tardigrade Admin · Accepted Answer

Substituting ln x = t, we get ∫ limits12010(1+(1-t/t(t-( ln t)))) d t=2009+∫ limits12010- ln 2010 (-1/u) d u (u=t- ln t) =2009- ln (2010- ln 2010)

Q. The value of the definite integral $e \int e^{2010} \frac{1}{x} (1 + \frac{1 - l n x}{l n x l n ( \frac{x}{l n x} )}) d x$ , equals

$2009 - ln (2010 - ln 2010)$

$2010 - ln (2009 - ln 2009)$

$2009 - ln (2010 - ln 2009)$

$2010 - ln (2010 - ln 2010)$

Substituting $ln x = t$ , we get
$1 \int 2010 (1 + \frac{1 - t}{t ( t - ( l n t ))}) d t = 2009 + 1 \int 2010 - l n 2010 \frac{- 1}{u} d u (u = t - ln t)$
$= 2009 - ln (2010 - ln 2010)$

Q. The value of the definite integral e∫e2010​x1​(1+lnxln(lnxx​)1−lnx​)dx, equals

2009−ln(2010−ln2010)

2010−ln(2009−ln2009)

2009−ln(2010−ln2009)

2010−ln(2010−ln2010)