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.

$$\int \frac{x^2}{x^2+2x+1}=\int\frac{x^2+2x+1}{x^2+2x+1}+\int \frac{-2x-2}{x^2+2x+1}+\int \frac{1}{x^2+2x+1} $$
Can you take it from here?
