The integral of $ \int_{0}^{1} 2^{x^2 +x} dx$ So, What I tried was,
$$ I(b) =  \int_{0}^{1} 2^{x^2 + x +b} dx$$
And hence,
$$ I'(b) = \ln(2)  I$$
Hence,
$$ I = C_o 2^{b}$$
or,
$$ C2^{b} = \int_{0}^{1} 2^{x^2 +x + b} dx$$
Now I'm trying to find an easy 'b' to evaluate the right side integral at, so as to figure out my constant. However I'm not sure how to find that 'b'. A guess was to take $ b= \frac{1}{4}$ however that was not a fruitful substitution

If you do not want to use special functions, build a series expansion around $x=\frac 12$ to get for the integrand
$$2^{3/4} \left(1+2 L \left(x-\frac{1}{2}\right)+L (2 L+1) \left(x-\frac{1}{2}\right)^2+\frac{2}{3}
   L^2 (2 L+3) \left(x-\frac{1}{2}\right)^3+\frac{1}{6} L^2 (4 L^2+12L+3)
   \left(x-\frac{1}{2}\right)^4+L^3\left(\frac{4 L^2}{15}+\frac{4 L}{3}+1\right)
   \left(x-\frac{1}{2}\right)^5+O\left(\left(x-\frac{1}{2}\right)^6\right) \right)$$ where $L=\log(2)$.
Integrate termwise and an approximation is
$$\int_{0}^{1} 2^{x^2 +x} dx \sim \frac{480+40 L+83 L^2+12 L^3+4 L^4 } {240 \sqrt[4]{2} }\approx 1.93589$$ while the exact solution is $1.93749$.
For sure, adding terms will improve the accuracy.
