Question

evluate :

Answer 1

i am writing it now

Answer 2

\[\int\limits_{1}^{e}(2\ln10logx)/(x) dx\]

Answer 3

You want to do a u-sub. Let u=log(x) since it is log base 10 the derivative is:
du=(1/x)(1/ln(10))dx
So:
ln(10)du=(1/x)dx
Giving:
\[2 (\ln(10))^2 \int\limits u du=2 \ln^2(10)\frac{u^2}{2}=\ln^2(10)\log^2(x)|^{e}_1=\ln^2(10)\log^2(e)\]
Then you can use the change of base to do:
\[\log^2(e)=(\frac{\ln(e)}{\ln(10)})^2=\frac{1}{\ln^2(10)}\]
Which makes your integral 1.

Answer 4

what would be the next step? could you show me

Answer 5

From where? The very last step I just replaced log^2(e) with 1/ln^2(10) and since you already have that:
ln^2(10)/ln^2(10)=1

Or was there anther step you were talking about?

Answer 6

so that would be the answer

Answer 7

yeah, just 1. :P

Answer 8

cool thanks

Answer 9

No problem :P