From "Mathamatical Analysis" of T.M. Apostol.
Define
\begin{gather*} f(x)=\left( \int_0^xe^{-t^2} \ dt\right)^2 \ \ \ \text{and} \ \ \ \ g(x)=\int_0^1 \frac{e^{-x^2(t^2+1)}}{t^2+1} \ dt \end{gather*}
a) Show that $g'(x) + f'(x) = 0$ for all $x$ and deduce that $g(x) + f(x) = \frac{\pi}4$.
I've computed the derivatives
\begin{gather*} f'(x)=2\int_0^xe^{-x^2-t^2} \ dt \ \ \ \text{and} \ \ \ \ g'(x)=-2x\int_0^1 e^{-x^2(t^2+1)} \ dt \end{gather*}
Now I think I need some change of variables to allow the summation of the integrands. Any help would be appreciated.
EDIT
With the hint in the comment below, using the change of variable $\mu(t)=xt$:
\begin{gather*} g'(x)=-2\int_0^1 e^{-x^2(t^2+1)} \ d xt = -2\int_0^x e^{-t^2-x^2} \ d t \end{gather*}
This is enough to show $f'(x)+g'(x)=0$.
To compute $f+g$, take $x=0$ so $f=0,\,g=\int_0^1\frac{dt}{t^2+1}=\frac{\pi}{4}$ (e.g. by $t=\tan u$).