Something wrong at $\int \frac{x^2}{x^2+2x+1}dx$ I have to calculate $$\int \frac{x^2}{x^2+2x+1}dx$$ and I obtain: $$\int \frac{x^2}{x^2+2x+1}dx=\frac{-x^2}{x+1}+2\left(x-\log\left(x+1\right)\right)$$
but I verify on wolfram and this is equal with:  $$x-\frac{1}{x+1}-2\log\left(x+1\right)$$ where did I go wrong?

P.S: Here is all steps: $$\int \:\frac{x^2}{x^2+2x+1}dx=-\frac{x^2}{x+1}+\int \:\frac{2x}{x+1}dx=-\frac{x^2}{x+1}+2\left(x-\log\left(x+1\right)\right)$$ I am sure that is not wrong, but the form on wolfram seems easily then it.

You are not wrong: your answer and the Wolfram answer differ by a constant. Both answers have a term of $-2\log(x+1)$. As for the rest:
$$
\text{(Yours) - (Wolfram)} = \left(2x - {x^2\over x+1}\right)
-
\left(x - {1\over x+1}\right)
=
x - {x^2-1\over x+1} = x - (x-1) = 1
$$
