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

Oct 2, 2017

$I = \left(x + 1\right) \ln \left(x + 1\right) - x + \setminus \frac{1}{2} {\ln}^{2} \left(x + 1\right) - \setminus \frac{{x}^{2}}{2}$

Explanation:

Starting with

$I = \setminus \int \left[\ln \left(x + 1\right) - \setminus \frac{\ln \left(x + 1\right)}{x + 1} - x\right] \mathrm{dx}$

I used $\setminus \int \ln \left(x\right) = x \ln x - x$
You can check this result by taking the derivative of both sides.
$\ln x = \ln x + 1 - 1 = \ln x$

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

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