What is #int ((x+1)*ln(x+1))/(x+2)-x#?

1 Answer
Oct 2, 2017

Answer:

#I = (x+1)ln(x+1) - x +\frac{1}{2}ln^2(x+1) - \frac{x^2}{2}#

Explanation:

Starting with

#I = \int [ln(x+1) - \frac{ln(x+1)}{x+1} - x] dx#

I used #\int ln(x) = xln x - x#
You can check this result by taking the derivative of both sides.
#lnx = lnx + 1 - 1 = lnx#

For the second term we see that #\frac{1}{x+1}# is the derivative of #ln(x+1)# and is thus of the form #f df/dx = \frac{1}{2}\frac{df^2}{dx}#

The third term is simply the integral of x and is #\frac{1}{2}x^2#