Answer
$f(x)=\displaystyle \frac{1}{8}(x^{2}+1)^{4}-\frac{1}{8}$
Work Step by Step
Slope of tangent line = $f'(x)=x(x^{2}+1)^{3}$
so,
$f(x)=\displaystyle \int x(x^{2}+1)^{3}dx=$
Substitute:
$\left[\begin{array}{ll}
u=x^{2}+1, & du=2xdx\\
& xdx=du/2
\end{array}\right]$
$f(x)=\displaystyle \frac{1}{2}\int u^{3}du\qquad$
apply the power rule:
$=\displaystyle \frac{1}{8}u^{4}+C$
$=\displaystyle \frac{1}{8}(x^{2}+1)^{4}+C.$
Given that $f(0)=0$, we find $C:$
$\displaystyle \frac{1}{8}(0+1)^{4}+C=0$
$C=-\displaystyle \frac{1}{8}$
Thus,
$f(x)=\displaystyle \frac{1}{8}(x^{2}+1)^{4}-\frac{1}{8}$
