Solving an Integral

Question

If $\displaystyle{g(x)=\int_0^1 \frac{e^{-x^2(t^2+1)}}{t^2+1}\,dt}$, find $g'(x)$. Any hint?

Answer

Let assume: $\int f(x, t)dt = F(x, t) + c_0$ $$g(x)=\int_{\phi(x)}^{\psi(x)}{f(x, t)}dt =\left|F(x, t)\right|_{t = \phi(x)}^{t = \psi(x)} = F(x, \psi(x)) - F(x, \phi(x))$$ \begin{align*} g'(x) &= \left[\frac {\partial}{\partial x}F(x, \psi(x)) + \frac {\partial}{\partial \psi(x)}F(x, \psi(x)) \cdot {\color{red}{\psi'(x)}}\right] - \left[\frac {\partial}{\partial x}F(x, \phi(x)) + \frac {\partial}{\partial \phi(x)}F(x, \phi(x)) \cdot {\color{red}{\phi'(x)}}\right] \\ &= \left[\frac {\partial}{\partial x}F(x, \psi(x)) - \frac {\partial}{\partial x}F(x, \phi(x))\right] \end{align*} $$\begin{align*}\color{green}g'(x) = \int_{\phi}^{\psi} \frac {\partial}{\partial x}f(x, t)dt {g'(x)} & = \int_0^1 {\frac{\partial}{\partial x}\frac{e^{-x^2(t^2+1)}}{t^2+1}}dt\\ & = -2x\int_0^1 {\frac{e^{-x^2(t^2+1)}}{t^2+1}}dt\\ & = -2x\int_0^1e^{-x^2t^2}.e^{{-x^2}}dt\\ & = -2xe^{{-x^2}}\int_0^1e^{-x^2t^2}dt\\ & = -2xe^{-x^2}\times \frac {\sqrt\pi}{2} \times \frac{\text{erf}(x)}{x}\\ & = \color{green}{-\sqrt\pi e^{-x^2} \text{erf}(x) = -e^{-x^2}\gamma(\frac 1{2}, x^2)} \end{align*}$$