∫ limits1e log x dx =

Question

∫ limits1e log x dx =  (A) 1 (B) e - 1 (C) e + 1 (D) 0

Tardigrade Admin · Accepted Answer

We have, ∫ limits1e log x dx = =[ log x ∫ limits 1 dx ]e1 - ∫ limits1e ( (d/dx) log x ∫ limits 1 dx ) dx =[ log x . x ]e1 - ∫ limits1e (1/x) x dx = [e log e - 1 log (1) ] - ∫ limits1e 1 dx = e - [x]e1 =e - [e - 1] = 1

Q. $1 \int e lo g x d x =$

$1$

$e - 1$

$e + 1$

$0$

We have, $1 \int e lo g x d x =$
$= [lo g x \int 1 d x]_{1}^{e} - 1 \int e (\frac{d}{d x} lo g x \int 1 d x) d x$
$= [lo g x . x]_{1}^{e} - 1 \int e \frac{1}{x} x d x$
$= [e lo g e - 1 lo g (1)] - 1 \int e 1 d x$
$= e - [x]_{1}^{e} = e - [e - 1] = 1$

Q. 1∫e​logxdx=

1

e−1

e+1

0

We have, 1∫e​logxdx= =[logx∫1dx]1e​−1∫e​(dxd​logx∫1dx)dx =[logx.x]1e​−1∫e​x1​xdx =[eloge−1log(1)]−1∫e​1dx =e−[x]1e​=e−[e−1]=1

Q. $1 \int e lo g x d x =$

$1$

$e - 1$

$e + 1$

$0$

We have, $1 \int e lo g x d x =$
$= [lo g x \int 1 d x]_{1}^{e} - 1 \int e (\frac{d}{d x} lo g x \int 1 d x) d x$
$= [lo g x . x]_{1}^{e} - 1 \int e \frac{1}{x} x d x$
$= [e lo g e - 1 lo g (1)] - 1 \int e 1 d x$
$= e - [x]_{1}^{e} = e - [e - 1] = 1$