Evaluate $\int_{0}^{1}2^{x^2+x}\mathrm dx$ 
Question : Evaluate - $$\int_{0}^{1}2^{x^2+x}\mathrm dx$$

My Attempt : First I tried to evaluate the indefinite integral of $2^{x^2+x}$ in order to put the limits $0$ and $1$ later on, but couldn't integrate it. Then I checked on WA and came to know that it's elementary integral doesn't exist. Now I moved one to using properties of definite integration such as $$\int_a^b f(x) \mathrm dx=\int_a^b f(a+b-x) \mathrm dx$$
But it couldn't help either. Can you please give me hint to proceed on this question?
P.S. - This is a high school level problem and therefore its solution shouldn't involve any special functions, such as Gaussian Integral etc.
Edit :  I asked my teacher this question and basically this was an approximation based question. This was a MCQ type question which has an option "None of the above" and it was the correct answer, since the other options were made in such a way that can be rejected by bounding this integral between 2 functions. For example we can use $$2^{x^2+x}<2^{2x} ~; ~x\in (0,1)$$ and thus can be sure that this integral is less than $3/\ln(4)$.
Thanks all for devoting your time in my question!
 A: Hint:
Put $y = 2^{x^2+x}$
Integral becomes $\int_{1}^{4} \frac{1}{\sqrt{1+ \frac{4}{\ln(2)}\ln(y)}}dy$
Again Put $\sqrt{1+ \frac{4}{\ln(2)}\ln(y)}= u$
Integral becomes $\int_{1}^{3}\frac{1}{2e^a} e^{au^2} du$
where $a = \frac{\ln(2)}{4}$
It resembles the standard integral $\int_{1}^{3} e^{au^2}du$
$$\int e^{au^2}du = \frac{-i\sqrt{\pi}}{2\sqrt{a}} \text{erf}\left(iu\sqrt{a}\right)$$
I hope you can take it from there
I am attaching the table of standard integrals for your reference
http://integral-table.com/downloads/integral-table.pdf
see page page 7, integral number 67
Good luck
A: If you cannot use special functions, then either numerical integration or approximation would be required.
For example, consider the Taylor expansion built around $x=\frac 12$ (mid point of the integration interval selected in order to tvoid promoting one of the bounds). You would  get 
$$2^{x^2+x}=2^{3/4}+2^{3/4} \left(x-\frac{1}{2}\right) \log (4)+2^{3/4}
    \log (2) (1+\log
   (4))\left(x-\frac{1}{2}\right)^2+O\left(\left(x-\frac{1}{2}\right)^3\right)$$ Integrate termwise to get 
$$\int 2^{x^2+x}\,dx=2^{3/4} \left(x-\frac{1}{2}\right)+\frac{\left(x-\frac{1}{2}\right)^2 \log
   (4)}{\sqrt[4]{2}}+\frac{1}{3} 2^{3/4} \left(x-\frac{1}{2}\right)^3 \log (2)
   (1+\log (4))+O\left(\left(x-\frac{1}{2}\right)^4\right)$$ USe the bounds to get, as an approximation,
$$\int_0^1 2^{x^2+x}\,dx\approx\frac{24+\log ^2(4)+\log (4)}{12 \sqrt[4]{2}}\approx 1.91361$$ while Wolfram Alpha would give $\approx 1.93749$.
For sure, you could improve using more terms. For illustration purposes, suppose that we make the expansion to $O\left(\left(x-\frac{1}{2}\right)^n\right)$. We should get
$$\left(
\begin{array}{cc}
n & \text{result} \\
 2 & 1.91361 \\
 4 & 1.93589 \\
 6 & 1.93741 \\
 8 & 1.93749
\end{array}
\right)$$
A: Hint:
Using $u=\frac{2x+1}{2}$ yields an imaginary error function.
