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