For all $x$ in $\mathbb R$ define $\displaystyle f(x)=\left(\int_0 ^{x} e^{-t^2}dt\right)^2$ and $\displaystyle g(x)=\int_{0}^{1}\frac{e^{-x^2(t^2+1)}}{t^2+1}dt$. Show that for all $x$ in $\mathbb R$ f'(x)+g'(x)=0
I did:
\displaystyle f'(x)=2\left( \int_{0}^{x}e^{-t^2}dt\right)e^{-x^2} and \displaystyle g'(x)=\int_{0}^{1}e^{-x^2(t^2+1)}(-2x)dt then changing $xt\rightarrow t$
\displaystyle g'(x)=-2x e^{-x^2}\int_{0}^{x}e^{t^2}dt , finally
\displaystyle f'(x)+g'(x)=2(1-x)e^{-x^2}\int_{0}^{x}e^{t^2}dt then this is equal to zero only if x=1.
Am i missing something? thanks beforehand.