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

Question

1864 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-02T16:23:18

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

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 $\frac{1}{2}x^2$

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