$\int \frac{\left(2x^3-4x^2-x-3\right)}{\left(x^2-2x-3\right)}dx$ confusion

Question

$$\int \frac{\left(2x^3-4x^2-x-3\right)}{\left(x^2-2x-3\right)}dx$$ =$$x^2+3\ln \left(x-3\right)+2\ln \left(x+1\right)+C$$

Where did the $x^2$ come from?

What I did:

Partial fraction then integrate, got the answer without $x^2$ term.

Answer 1

Hint. It comes from the partial fraction decomposition, you have $$ \frac{2x^3-4x^2-x-3}{x^2-2x-3}=2x+\frac2{x+1}-\frac3{x-3} \tag1 $$ giving $$ \begin{align} \int\frac{2x^3-4x^2-x-3}{x^2-2x-3}dx&=\int 2xdx+\int\frac2{x+1}dx-\int\frac3{x-3}dx\\\\ &=\color{red}{x^2}+\int\frac2{x+1}dx-\int\frac3{x-3}dx\\\\ &=\color{red}{x^2}+2 \ln|x+1|-3 \ln|x-3|+C. \end{align} $$

Answer 2

$$\int \frac{\left(2x^3-4x^2-x-3\right)}{\left(x^2-2x-3\right)}dx$$

$$=\int\bigg(2x+\frac{3}{x-3}+\frac{2}{x+1}\bigg)dx$$

$$=2\int\frac{1}{x+1}dx+3\int\frac{1}{x-3}dx+2\int xdx$$

$$\boxed{\color{red}{=\color{blue}{x^2}+2 \ln|x+1|-3 \ln|x-3|+C}}$$