Integration of some weird function I am just wondering, is there a closed form for such integral? I am thinking along the lines of polar substitution but I cannot exactly figure it out. Any help will be greatly appreciated. Thanks. 
$$\int_0^1\frac{1}{\sqrt{x^2+(1-x)^2}}  e^{\large{-\frac{z^2}{2[{x^2+(1-x)^2]}}}}\ \mathrm{d}x$$
*[Additional] Now, I tried this (as commented by User45878). 
Suppose I let $$f(z)=\int_0^1\frac{1}{\sqrt{x^2+(1-x)^2}}  e^{\large{-\frac{z^2}{2[{x^2+(1-x)^2]}}}}\ \mathrm{d}x$$ From which I get
$$\frac{\partial}{\partial z}f(z)=\frac{-z}{x^2+(1-x)^2}f(z)$$
So that 
$$\frac{\frac{\partial}{\partial z}f(z)}{f(z)}=\frac{-z}{x^2+(1-x)^2}$$
Am I going on the right track, or it will just lead me nowhere? I am not so sure how to apply the "differentiation under integration" technique.
 A: By noticing that $2\left(x^2+(1-x)^2\right)=(2x-1)^2+1$ we have:
$$I_z=\int_0^1\frac{1}{\sqrt{x^2+(1-x)^2}}  e^{\large{-\frac{z^2}{2[{x^2+(1-x)^2]}}}}\,dx = \sqrt{2}\int_{0}^{1}\frac{1}{\sqrt{1+y^2}}\exp\left(-\frac{z^2}{1+y^2}\right)\,dy,$$
$$I_z=\frac{1}{\sqrt{2}}\int_{1/2}^{1}\frac{1}{u\sqrt{(1-u)}}e^{-z^2u}\,du=\sqrt{2}\int_{1/\sqrt{2}}^{1}\frac{1}{u\sqrt{1-u^2}}e^{-z^2u^2}\,du,$$
$$I_z = \sqrt{2}\int_{\pi/4}^{\pi/2}\frac{\exp\left(-z^2\sin^2\theta\right)}{\sin\theta}\,d\theta=\sqrt{2}\int_{0}^{\pi/4}\frac{\exp\left(-z^2\cos^2\theta\right)}{\cos\theta}\,d\theta.\tag{1}$$
By differentiating with respect to $z$ we get:
$$\frac{dI_z}{dz}=-2\sqrt{2}\,z\cdot\int_{0}^{\pi/4}\cos\theta\cdot\exp\left(-z^2\cos^2\theta\right)\,d\theta,$$
$$\frac{dI_z}{dz}=-2\sqrt{2}\,z\,e^{-z^2}\cdot\int_{0}^{\pi/4}\cos\theta\cdot\exp\left(z^2\sin^2\theta\right)\,d\theta.\tag{2}$$
By taking the Taylor series of the exponential function we can write:
$$\frac{dI_z}{dz}=-2\sqrt{2}\,z\,e^{-z^2}\cdot\sum_{k=0}^{+\infty}\frac{z^{2k}}{k!}\int_{0}^{\pi/4}\cos\theta\sin^{2k}\theta\,d\theta,$$
$$\frac{dI_z}{dz}=-2\,e^{-z^2}\cdot\sum_{k=0}^{+\infty}\frac{z^{2k+1}}{2^{k}(2k+1)k!}=-2e^{-z^2}\cdot\int_{0}^{z}e^{w^2/2}dw.\tag{3}$$
Since a rescaled Dawson integral appears in the RHS, $I_z$ does not have a closed form in the common sense. Anyway, by taking derivatives in $(3)$ we get that $I_z$ satisfies the differential equations:
$$\frac{d^2 I_z}{dz^2}+2z\frac{dI_z}{dz}+2e^{-z^2/2}=0,\tag{4}$$
$$\frac{d^3 I_z}{dz^3}+(2+2z^2)\frac{d^2 I_z}{dz^2}+(2+z)\frac{dI_z}{dz}=0.\tag{5}$$
By $(1)$, an expression for $I_z$ is given by:
$$I_z=\sqrt{2}\,e^{-z^2}\sum_{k=0}^{+\infty}\frac{z^{2k}}{k!}\int_{0}^{\pi/4}\frac{\sin^{2k}(\theta)}{\cos(\theta)}d\theta = \sqrt{2}\,e^{-z^2}\sum_{k=0}^{+\infty}\frac{z^{2k}}{k!}\int_{0}^{1/\sqrt{2}}\frac{u^{2k}}{1-u^2}du,\tag{6}$$
where:
$$\int_{0}^{1/\sqrt{2}}\frac{u^{2k}}{1-u^2}du=\int_{0}^{1/\sqrt{2}}\left(\frac{1}{1-u^2}-\frac{1-u^{2k}}{1-u^2}\right)du,$$
$$\int_{0}^{1/\sqrt{2}}\frac{u^{2k}}{1-u^2}du=\log(1+\sqrt{2})-\sqrt{2}\sum_{j=1}^{k}\frac{1}{2^j(2j-1)}=\sqrt{2}\sum_{j>k}\frac{1}{2^j(2j-1)}.\tag{7}$$
By writing $e^{-z^2}$ as a Taylor series and by regarding $(6)$ as a Cauchy product we get:
$$I_z = 2\sum_{n=0}^{+\infty}\frac{z^{2n}}{n!}\int_{0}^{1/\sqrt{2}}(1-u^2)^{2n-1}du,\tag{8}$$
in which we see values of the incomplete Beta function, as already pointed by user121049.
A: Could do a power series in z. The coefficients are nasty looking incomplete beta functions (courtesy of Wolfram Alpha), but maybe they simplify.
