Question

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

Answer 1

`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`