Evaluate the integral [ln(x)/(x) dx]

Tuesday, February 8, 2011

Evaluate the integral [ln(x)/(x) dx]

Evaluate $\displaystyle\int$
(lnx)/x dx $\displaystyle:$

We let $\displaystyle{u}={\ln{{x}}}$ . Then $\displaystyle{d}{u}=\frac{{1}}{{x}}{\left.{d}{x}\right.}$ and we have:

$\displaystyle\int{u}{d}{u}=\frac{{1}}{{2}}{{u}}^{{2}}+{C}$ . Substituting for $\displaystyle{u}$ we get $\displaystyle\frac{{1}}{{2}}{{\left({\ln{{x}}}\right)}}^{{2}}+{C}$ .

Thus the integral evaluates as $\displaystyle\frac{{1}}{{2}}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}+{C}$

href="https://en.wikipedia.org/wiki/Logarithm">https://en.wikipedia.org/wiki/Logarithm

Spooky Notes

Tuesday, February 8, 2011

Evaluate the integral [ln(x)/(x) dx]

No comments:

Post a Comment

How is Joe McCarthy related to the play The Crucible?