Answer
$$
f(x)=\frac{2 x^{2}+5}{x^{2}+1}
$$
The general anti-derivative of the function $f(x)$ is given by:
$$
\begin{aligned}
F(x) &= \int f(x)dx \\
&=2 x+3 \tan ^{-1} x+C
\end{aligned}
$$
where $C$ is an arbitrary constant.
Work Step by Step
$$
f(x)=\frac{2 x^{2}+5}{x^{2}+1}
$$
To find general antiderivative of the function $f(x) $ . First rewrite $f(x) $ as follows :
$$
f(x)=\frac{2 x^{2}+5}{x^{2}+1}=\frac{2\left(x^{2}+1\right)+3}{x^{2}+1}=2+\frac{3}{x^{2}+1}
$$
So, the general anti-derivative of the function $f(x)$ is given by:
$$
\begin{aligned}
F(x) &= \int f(x)dx \\
&=\int \left( 2+\frac{3}{x^{2}+1}\right) dx\\
&=2 x+3 \tan ^{-1} x+C
\end{aligned}
$$
where $C$ is an arbitrary constant.
