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
 A: Hint:
$$\int{2^{x^{2} + x} d x} = \int{e^{\left(x^{2} + x\right) \ln{\left(2 \right)}} dx}$$
Completing the square,
$$\int{\frac{2^{\frac{3}{4}} e^{\left(x \sqrt{\ln{\left(2 \right)}} + \frac{\sqrt{\ln{\left(2 \right)}}}{2}\right)^{2}}}{2} d x}$$
If
$$t=x \sqrt{\ln{\left(2 \right)}} + \frac{\sqrt{\ln{\left(2 \right)}}}{2} \to dx = \frac{dt}{\sqrt{\ln{\left(2 \right)}}}$$
Hence,
$$\int{\frac{2^{\frac{3}{4}} e^{t^{2}}}{2 \sqrt{\ln{\left(2 \right)}}} d t} = \frac{2^{\frac{3}{4}} \displaystyle\int{e^{t^{2}} d t}}{2 \sqrt{\ln{\left(2 \right)}}}$$
and the integral of $e^{t^{2}}$does not have a closed form, and occour to use imaginary error function called $\text{erfi}$. After you can calculate the definite integral.
A: 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.
A: Here is a (failed) attempt using the Leibniz rule (a.k.a. Feynman's trick).
First changee variables (Ninad comment);  the problem is equaivalent to
evaluating
$$
\int_{1/4}^{3/4} e^{(\log 2) x^2}dx .
$$
Let
$$
F(b) = \int_{1/4}^{3/4} e^{b x^2}dx
$$
We want to evaluate $F(\log 2)$.
Differentiate (using the Leibniz rule)
$$
F'(b) = \int_{1/4}^{3/4} x^2e^{b x^2}dx
$$
Integrate by parts
$$
F'(b) = \frac{3e^{9b/4} - e^{b/4}}{4b} - \frac{1}{b^2}\int_{1/4}^{3/4}e^{b x^2}dx
$$
So $F$ satisfies the differential equation
$$
F'(b)+\frac{1}{b^2}F(b) = \frac{3e^{9b/4} - e^{b/4}}{4b} .
\tag{1}$$
Now the homogeneous equation
$$
G'(b)+\frac{1}{b^2}G(b) = 0
$$
is easily solved, $G(b) = Ce^{1/b}$, the inhomogeneous
DE $(1)$ is not so easy.  The solution
$e^{1/b}$ is not of the proper form in order to use undetermined coefficients.  We can always try variation of parameters; but there are integrals in there, and they turn out to be just as hard as the original problem!
$$
F(b) = Ce^{1/b} + e^{1/b}
\int \frac{3 e^{(9b^2-4)/(4b)}-e^{(b^2-4)/(4b)}}{4b} \;db
$$
