Evaluating the indefinite integral $\int e^{-x^2}\,\mathrm{d}x$ In my book, it is said that $$\int e^{-x^2} \, \mathrm{d}x$$ cannot be solved by the method of inspection. It then turned to method of substitution as a new topic. I am not able to solve this expression by method of substitution either. Can someone help me? Thanks.
 A: This is not an integral with a nice form. 
Note that this doesn't mean that all definite versions of this integral don't have nice solutions: for example, $$\int_{-\infty}^\infty e^{-x^2}dx=\sqrt{\pi};$$ this is known as the Gaussian integral.
See https://en.wikipedia.org/wiki/Gaussian_integral for further info.
A: This antiderivative is not solvable by any of the methods of basic calculus, including the method of substitution. This is proven using the methods of a field of math called differential Galois Theory. However, that doesn't stop mathematicians from inventing a function to solve this. This function is the error function, denoted $\text{erf}$, and is applied in this way:
$$\int e^{-x^2}dx=\frac{\sqrt{\pi}}{2}\text{erf}(x).$$
When I first learned this, I was highly dissatisfied with this answer, but this is (depending on who you ask) the definition of the error function, and its sole purpose is to solve this type of integral.
A: The integral does not have an elementary anti derivative. That is, it can't be expressed in terms of "simple" functions such as logarithms, exponentials, trigonometrical functions, etc... We have, however that:
$$\int e^{-x^2}\, \mathrm{d}x=\frac{\sqrt{\pi}}{2}\text{erf}(x)$$
where $\text{erf}(x)$ is the error function. We also have a quite a few nice definite integrals, such as the Gaussian Integral:
$$\int_{-\infty}^{\infty} e^{-x^2}\, \mathrm{d}x=\sqrt{\pi}$$
And since the integral is symmetric, we have also 
$$\int_{0}^{\infty} e^{-x^2}\, \mathrm{d}x=\frac{\sqrt{\pi}}{2}$$ 
A: The integral in question does not have a closed form expression in terms of elementary functions. You could expand the integral into $$\int e^{-x^2} {dx}=\int \left(\sum_{k=0}^{\infty} \frac{(-1)^k x^{2k}}{k!}\right) dx$$ And from there obtain a power-series representation of the indefinite integral. This also has the benefit of being convergent for all real numbers $x$. The first few terms in the series representation of the integral are$$C+x-\frac{x^3}{3}+\frac{x^5}{10}-\frac{x^7}{42}...$$Where C is an arbitrary constant of integration.
A: $e^{-x^2}$ does not have an elementary antiderivative. In other words, there is no 'simple' function that this integral is equal to.
The integral 
$$\int_{-\infty}^{\infty}e^{-x^2}\ dx$$
on the contrary, can be evaluated by polar coordinates with double integrals.
A: This integral is definitively the Error function and no it cannot be evaluated by using substitution. However, by setting certain limits, for instance
\begin{equation}I=\int_0^{\infty}e^{-x^2}dx
\end{equation}
the integral can be solved (by means of the technique called differentiation under the integral sign) to achieve special values.
A: This is indeed a very poor choice for an illustration, as no substitution can help in this case (as other respondents say, this antiderivative is hopeless.)
A better example could be
$$\int xe^{-x^2}\,dx.$$
If you substitute $t=-x^2$, you have that $dt=-2x\,dx$ and you can write
$$\int xe^{-x^2}\,dx=-\frac12\int e^t\,dt.$$
This is an easy antiderivative that gives
$$-\frac{e^t}2+C=-\frac{e^{-x^2}}2+C.$$
