Calculus Problem of the Week: June 13, 2011

Starting this week, we will start posting a calculus problem with its solution (hidden of course!)! Sometimes easy and sometimes hard, these problems could come from any calculus topic. If you really want to get better at calculus, following these problems is a great way to make yourself practice!

This week’s problem
Find the integral: \int\!\dfrac{\ln(x)}{x}\,\mathrm{d}x

See the solution.

This problem can be done with a straightforward u substitution. If you let u=\ln(x) then \mathrm{d}u=\dfrac{1}{x}\mathrm{d}x. Solving for \mathrm{d}x then gives \mathrm{d}x = x\mathrm{d}u.

Therefore, our integral becomes:
\int\!\dfrac{\ln(x)}{x}\,\mathrm{d}x=\int\!\dfrac{u}{x}\,x\mathrm{d}u=\int\!u\,\mathrm{d}u=\dfrac{u^2}{2}+C=\dfrac{(\ln(x))^2}{2}+C

Hide this solution.