What is the Integral of #(x+1)/x dx#?

1 Answer
Jun 4, 2015

Not many people realize this, but the key to doing this is to separate it using the additive properties of integrals.

#int (x+1)/xdx#

#= int cancel(x/x)dx + int 1/xdx#

#= int dx + int 1/xdx#

#= color(blue)(x + ln|x| + C)#

where you'll have to remember that the integral of #1/u((du)/(dx))# is #ln|u|#. You'll see it a lot more later.