We have,
$$
\begin{gathered}
\int x(1+x) \log \left(1+x^2\right) d x=f(x) \log \left(1+x^2\right)- \\
\frac{2}{3} \tan ^{-1} x-\frac{2 x^3}{9}-\frac{x^2}{2}+\frac{2 x}{3}+c \ldots .
\end{gathered}
$$
Let $\quad I=\int x(1+x) \log \left(1+x^2\right) d x$
$$
=\int\left(x+x^2\right) \log \left(1+x^2\right) d x
$$
Now, by using integration by parts, we get
$$
\begin{gathered}
I=\log \left(1+x^2\right) \cdot\left(\frac{x^2}{2}+\frac{x^3}{3}\right)-\int \frac{1}{1+x^2} \cdot \\
2 x \cdot\left(\frac{x^2}{2}+\frac{x^3}{3}\right) d x \\
=\log \left(1+x^2\right) \cdot\left(\frac{x^2}{2}+\frac{x^3}{3}\right)-\int \frac{x^3}{1+x^2} d x \\
-\frac{2}{3} \int \frac{x^4}{1+x^2} d x
\end{gathered}
$$
Let $y_1=\int \frac{x \cdot x^2}{1+x^2} d x$
Put $x^2=t$


$$
\begin{aligned}
& \Rightarrow 2 x d x=d t \Rightarrow x d x=\frac{d t}{2} \\
& \therefore I_1=\frac{1}{2} \int \frac{t}{1+t} d t \\
& \quad=\frac{1}{2} \int \frac{1+t-1}{1+t} d t=\frac{1}{2} \int d t-\frac{1}{2} \int \frac{d t}{1+t} \\
& \quad=\frac{1}{2} t-\frac{1}{2} \log |1+t|+c_1 \\
& \quad=\frac{1}{2} x^2-\frac{1}{2} \log \left(1+x^2\right)+c_1
\end{aligned}
$$
Let $\quad I_2=\int \frac{x^4}{1+x^2} d x=\int \frac{x^4-1+1}{1+x^2} d x$
$$
\begin{aligned}
& =\int\left(\frac{x^4-1}{1+x^2}+\frac{1}{1+x^2}\right) d x \\
& =\int\left(\left(x^2-1\right)+\frac{1}{1+x^2}\right) d x \\
& =\frac{x^3}{3}-x+\tan ^{-1} x+c_2
\end{aligned}
$$
On substituting the value of $I_1$ and $I_2$ in Eq. (ii), we get
$$
I=\log \left(1+x^2\right)\left(\frac{x^2}{2}+\frac{x^3}{3}\right)-\frac{1}{2} x^2+\frac{1}{2} \log \left(1+x^2\right)
$$


$$
-\frac{2}{9} x^3+\frac{2}{3} x-\frac{2}{3} \tan ^{-1} x+c
$$


where, $C=\frac{-2}{3} C_2-C_1$
$$
\begin{gathered}
=\operatorname{Iog}\left(1+x^2\right)\left(\frac{x^2}{2}+\frac{x^3}{3}+\frac{1}{2}\right)-\frac{2}{3} \tan ^{-1} x \\
-\frac{2}{9} x^3-\frac{x^2}{2}+\frac{2}{3} x+C
\end{gathered}
$$
On comparing this with Eq. (i), we get
$$
f(x)=\frac{x^2}{2}+\frac{x^3}{3}+\frac{1}{2}
$$