<s>[INST] How to draw the graph of $f(x)=\int_0^x\left(\frac{t^3-2t^2-4}{t^2+1}\right)\ dt$ using only your calculator? $$f(x)=\int_0^x\left(\frac{t^3-2t^2-4}{t^2+1}\right)\ dt$$
I need to find the $x$ and $y$ intercepts, and the inflection points of the function $f(x)$ (with both $x$ and $y$ coordinates). I need to find it through the calculator and explain my answer. How do I find the antiderivative?
 [/INST] Hint. You may observe that, by partial fraction decomposition, you have
$$
\frac{t^3-2t^2-4}{t^2+1}=t-2-\frac{t+2}{t^2+1}
$$ giving, for the antiderivative that vanishes at $x=0$:
$$
\begin{align}
f(x)=\int_0^x\frac{t^3-2t^2-4}{t^2+1}dt&=\int_0^x\left(t-2-\frac{t+2}{t^2+1}\right)dt\\\\
&=\int_0^x\left(t-2\right)dt-\int_0^x\frac{t}{t^2+1}dt-2\int_0^x\frac{1}{t^2+1}dt\\\\
&=\frac{x^2}2-2x-\frac12\ln(x^2+1)- 2\arctan x.
\end{align}
$$
Remark. Here the constant of integration is equal to $0$ since $ \int_0^x$ is equal to $0$ when $x=0$.
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